4.22 Discrete Uniform Distribution

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The theoretical (expected) frequencies are

$\begin{array}{l} N P (0) & = (120) (0.22313) = 27, \\ N P (1) & = (120) (0.334695) = 40, \\ N P (2) & = (120) (0.25) = 30, \\ N P (3) & = (120) (0.13) = 16, \\ N P (4) & = (120) (0.05) = 6, \\ N P (5) & = (120) (0.01) = 1 \end{array}$ $\begin{array}{l} N P (0) & = (120) (0.22313) = 27, \\ N P (1) & = (120) (0.334695) = 40, \\ N P (2) & = (120) (0.25) = 30, \\ N P (3) & = (120) (0.13) = 16, \\ N P (4) & = (120) (0.05) = 6, \\ N P (5) & = (120) (0.01) = 1 \end{array}$

i.e., 27, 40, 30, 16, 6, and 1.

Example 4.73: If X and Y are two independent random Poisson random variables such that Var[X]+Var[Y]=3. Find P(X+Y<2)?

Solution: Let X and Y follow Poisson distribution.

Let λ₁ be the mean of X and λ₂ be the mean of Y.

Then by additive property X+Y follows Poisson distribution with mean λ₁+λ₂.

We have Var[X]+Var[Y]=λ₁+λ₂=3(given)

∴ λ=λ₁+λ₂=3 (say)

Now

$\begin{array}{l} P (X + Y < 2) = P (X + Y = 0) + P (X + Y = 1) & = e^{- λ} + \frac{e^{- λ} λ}{1!} \\ = e^{- 3 +} e^{- 3} 0.3 = e^{- 3} (1 + 3) \\ = 4 e^{- 3} = 0.199 \end{array}$ $\begin{array}{l} P (X + Y < 2) = P (X + Y = 0) + P (X + Y = 1) & = e^{- λ} + \frac{e^{- λ} λ}{1!} \\ = e^{- 3 +} e^{- 3} 0.3 = e^{- 3} (1 + 3) \\ = 4 e^{- 3} = 0.199 \end{array}$

Example 4.74: In a company, a board receives 15 calls per minute. If the board is capable of handling at the most 25 calls per minute, what is the probability that the board will saturate in 1 minute?

Solution: Number of calls received=15/minute.

∴ λ=15

The board reaches the saturation point if the number of calls exceeds 25.

$\begin{array}{l} Hence the probability of saturation & = 1 - \sum_{x = 0}^{25} \frac{e^{- 15} {(15)}^{x}}{x!} \\ = 1 - [e^{- 15} + \frac{e^{- 15} \cdot 15}{1!} + \dots + \frac{e^{- 15} {(15)}^{25}}{25!}] \\ = 1 - 0.994 = 0.006 \end{array}$ $\begin{array}{l} Hence the probability of saturation & = 1 - \sum_{x = 0}^{25} \frac{e^{- 15} {(15)}^{x}}{x!} \\ = 1 - [e^{- 15} + \frac{e^{- 15} \cdot 15}{1!} + \dots + \frac{e^{- 15} {(15)}^{25}}{25!}] \\ = 1 - 0.994 = 0.006 \end{array}$

Example 4.75: If V=Hourly volume=37, t=length of interval=30 seconds, then compute F_x for x>2.

Solution: We have

$λ = \frac{V t}{3600} = \frac{37 \times 30}{3600} = 0.308$ $λ = \frac{V t}{3600} = \frac{37 \times 30}{3600} = 0.308$

Therefore $e^{- λ} = e^{- 0.308} = 0.735, n = \frac{3600}{t} = \frac{3600}{30} = 120$ $e^{- λ} = e^{- 0.308} = 0.735, n = \frac{3600}{t} = \frac{3600}{30} = 120$

$\begin{array}{l} F_{0} & = n \frac{λ^{0} e^{- λ}}{0!} = 120 . e^{0.308} = 120 \times 0.735 = 88.2 \\ F_{1} & = n \frac{λ^{1} e^{- λ}}{1!} = \frac{λ}{1} F_{0} = 0.308 \times 88.2 = 27.2 \\ F_{2} & = n \frac{λ^{2} e^{- λ}}{2!} = \frac{λ}{2} F_{1} = \frac{0.308 \times 27.2}{2} = 4.2 \end{array}$ $\begin{array}{l} F_{0} & = n \frac{λ^{0} e^{- λ}}{0!} = 120 . e^{0.308} = 120 \times 0.735 = 88.2 \\ F_{1} & = n \frac{λ^{1} e^{- λ}}{1!} = \frac{λ}{1} F_{0} = 0.308 \times 88.2 = 27.2 \\ F_{2} & = n \frac{λ^{2} e^{- λ}}{2!} = \frac{λ}{2} F_{1} = \frac{0.308 \times 27.2}{2} = 4.2 \end{array}$

Hence F_x>2=n−(F₀+F₁+F₂)=120−(88.2+27.2+4.2)=0.4

Example 4.76: (One of the first published example of Poisson distribution as applied to traffic data by Adams). The rate of arrivals, i.e., the number of vehicles arriving per 10 second interval is given below. Obtain the theoretical frequencies, and hourly volume from the data.

No. of vehicles per 10 second period	0	1	2	3	>3
Observed frequency	94	63	21	2	0

Solution:

No. of vehicles per 10 second period (x)	Observed frequency (f)	Total of vehicles (xf)	Probability P(x)	Theoretical frequency
0	94	0	0.539	97.0
1	63	63	0.333	59.9
2	21	42	0.103	18.5
3	2	6	0.021	3.8
>3	0	0	0.004	0.8
	180	111	1.000	180

We have

$\begin{array}{l} λ = \frac{Total of vehicles}{Σ f} = \frac{111}{180} = 0.617 \\ e^{- λ} = e^{- 0.617} = 0.539 \end{array}$ $\begin{array}{l} λ = \frac{Total of vehicles}{Σ f} = \frac{111}{180} = 0.617 \\ e^{- λ} = e^{- 0.617} = 0.539 \end{array}$

Applying the formula $P (x) = \frac{λ^{x} e^{- x}}{x!}$ $P (x) = \frac{λ^{x} e^{- x}}{x!}$ we get

P(0)=0.539, P(1)=0.333, P(2)=0.103, P(3)=0.021, P(>3)=0.004

The theoretical frequencies are N P(0)=97, N P(1)=59.9, N P(2)=18.5, N P(3)=3.8, N P(X>3)=0.8

where N=180.

The hourly volume=2×no. of vehicles in 180 ten second period=2×111=222.

Exercise 4.8

1. If a random variable X follows Poisson distribution such that P(X=1)=P(X=2). Find (1) Mean, (2) P(X=0)?

Ans: P(X=0)=0.1353

2. A hospital receives patients at the rate of 3 patients/minute on an average. What is the probability of receiving no patient in 1 minute interval?

Ans: 0.04979

3. Suppose a book of 285 pages contains typographical errors. If these errors are randomly distributed throughout the book, what is the probability that 10 pages, selected at random will be free from errors (e^−0.735=0.4975).

Ans: 0.4795

4. In a Poisson distribution if 3P(X=2)=P(X=4). Find P(X=3)?

Ans: 36e⁻⁶

5. Between the hours 2 p.m. and 4 p.m. the average number of phone calls per minute coming into the switch board of a company is 2.35. Find the probability that during one particular minute there will be at most two phone calls?

Ans: 0.58285

6. Out of 1000 balls, 50 are red and the rest are white. If 60 balls are picked at random, what is the probability of picking up (a) 3 red balls, (b) not more than 3 red balls in the sample? Assume Poisson distribution for the number of red balls picked up in the sample where e⁻³=0.0498.

Ans: (a) 0.2241; (b) 0.6474

7. A manufacturer, who produces medicine bottles, finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 boxes from how many boxes will contain (1) no defectives (2) at least 2 defectives (given e^−0.5=0.60650).

Ans: (a) 61; (b) 9

8. The probability that a man aged 50 years will die within a year is 0.01125. What is the probability that out of 12 such men at least 11 will reach their 51st birthday (e^−0.135=0.87371)?

Ans: 0.9917

9. A pair of dice is thrown 6 times. If getting a total of 7 is considered to be a success, what is the probability of at least 4 successes?

Ans: $\frac{406}{46650}$ $\frac{406}{46650}$

10. There are 20% chances for a worker of an industry to suffer from an occupational disease. Fifty workers were selected at random and examined for the occupational disease. Find the probability that (a) only one worker is found suffering from the disease, (b) more than 3 are suffering from the disease, (c) None is suffering from the disease?

Ans: (a) $10 {(\frac{4}{5})}^{49}$ $10 {(\frac{4}{5})}^{49}$ ; (b) $1 - (\frac{4^{47}}{550}) (25364)$ $1 - (\frac{4^{47}}{550}) (25364)$ ; (c) ${(\frac{4}{5})}^{50}$ ${(\frac{4}{5})}^{50}$

11. If a random variable X follows a Poisson distribution such that P(X=2)=9 P(X=4)+90 P(X=6). Find the mean and variance of X?

Ans: Mean=21, Variance=1

12. A random variable has a Poisson distribution such that P(1)=P(2). Find the (a) Mean of the distribution, (b) P(4), (c) P(X≥1), (d) P(1<X<4)?

Ans: (a) 1.2; (b) 0.0902; (c) 0.8647; (d) 0.4511

13. Fit a Poisson distribution for the following data and calculate the expected frequencies.

x	0	1	2	3	4
f(x)	109	65	22	3	1

Ans: 108, 67, 66, 20, 4, 11, 1

14. A distributor of bean seeds determines from extensive tests that 5% of large batch of seeds will not germinate. He sells the seeds in packets of 200 and generates 90% germination. Determine the probability that a particular packet will violate the guarantee.

Ans: 0.0016

15. The number of accidents in a year to the taxi drivers in a city follows a Poisson distribution with mean 3. Out of 1000 taxi drivers, find approximately the number of drivers with (a) no accidents in a year, (b) more than 3 accidents in a year.

Ans: (a) 50; (b) 350

16. If the mean of a Poisson distribution is 4, find (a) SD, (b) β₁, (c) β₂, (d) μ₃, (e) μ₄.

Ans: (a) 2; (b) 0.25; (c) 3.25; (d) 4; (e) 52

17. One-hundred car radios are inspected as they come off the production line and the number of defects per set is recorded below:

No. of defects	0	1	2	3	4
No. of sets	79	18	2	1	0

Find the Poisson distribution to the above data and calculate the expected frequencies of 0, 1, 2, 3, and 4 defects?

Ans: 78, 20, 2, 0, 0

18. A book contains 100 misprints distributed randomly throughout its 100 pages. What is the probability that a page observed at random contains at least two misprints? Assume Poisson distribution.

Ans: 0.264

19. If X and Y are independent Poisson random variables with means 2 and 4, respectively, find (a) $P [\frac{X + Y}{2} < 1]$ $P [\frac{X + Y}{2} < 1]$ , (b) P[3(X+Y)≥9]?

Ans: (a) 0.017352; (b) 0.93803

20. For a Poisson distribution P(X=1)=0.03 and P(X=2)=0.2, find (a) P(X=0) and (b) P(X=3)?

Ans: (a) 0.00225; (b) 0.888899

21. Fit a Poisson distribution to the following data which gives the number of yeast cells per square for 400 squares.

No. of cells per square	0	1	2	3	4	5	6	7	8	9	10
No. of squares	103	143	98	42	8	4	2	0	0	0	0

Ans: 107, 14, 93, 41, 14, 4, 0, 0, 0, 0

22. The number of items collected at a center is Poisson distribution, where the probability of no items collected is 0.4. Find the probability that 3 or less items are collected?

Ans: 0.9856

23. A shopkeeper receives 5 bad currencies every day. What are the probabilities that he will receive (a) 3 bad currencies on a particular day, (b) 8 bad currencies for 2 consecutive days?

Ans: 0.1113

24. The following table compares the predicted frequencies with frequencies observed in a field study:

No. of vehicles arriving per interval	0	1	2	3
Observed frequency	88	27	5	0

Calculate the predicted frequencies.

Ans: 88.2, 27.2, 4.2, 0.4

25. Fit Poisson distribution by calculating predicted (theoretical) frequencies from the table given below:

No. of vehicles per 10 second period	0	1	2	>2
Observed frequency	88	27	5	0

Ans: 88.2, 27.2, 4.2, 0.4

26. Suppose that the chance of an individual coal miner is killed in a mining accident during a year is 1/1400. Use the Poisson distribution to calculate the probability that in the mine employing 350 minors, there will be at least one fatal accident.

Ans: 0.22

27. A car hire firm has two cars, which it hires day by day. The number of demands for a car on each day is distributed as a Poisson distribution with mean 1.5. Calculate the proportion of days on which no car is used and proportion of days on which some demand is refused.

Ans: 0.1913

28. Between the hours 2 p.m. and 4 p.m., the average number of phone calls per minute coming into the switch board is 2.35. Find the probability that during one particular unit there will be at most 2 phone calls?

Ans: 0.582854

29. The quality control manager of a tyre company has a sample of 100 tyres and has found the lifetime to be 30.214 km. The population SD is 860. Construct a 95% confidence interval for the mean life time for this particular brand of tyres.

Ans: (30045. 44, 303 82 .56)

30. In a random selection of 64 of 600 road crossings in a town, the mean number of automobile accidents per year was found to be 4.2 and the sample SD was 0.8. Construct a 95% confidence interval for the mean number of automobile accidents per crossing per year.

Ans: 4.023, 4.377

31. On a certain day 74 trains were arriving on time at Delhi station during the rush hours and 83 were late. At New Delhi there were 65 on time and 107 were late. Is there any difference in the proportions arriving on time at the two stations?

Ans: Null hypothesis (H₀): There is no difference of trains arriving “on time” at two stations is accepted.

32. The number of accidents in a year attributed to the state transport bus driver follows Poisson distribution with mean 3. Out of 1000 bus drivers, how many drivers had more than 1 accident in a year?

Ans: 801

33. For 10,000 insured cars of an insurance company, the average number of claims per year is said to be 2000. Using Poisson distribution, find the probability that there is 1 claim in a particular year?

Ans: 0.1638

34. A train runs 25 km at a speed of 30 km/h, another 50 km at a speed of 40 km/h, then due to repairs of the track it travels for 6 minutes at a speed of 10 km/h and finally covers the remaining distance of 24 km at a speed of 24 km/h. What is the average speed in kilometers per hour?

Ans: 31.41 km/h

4.22 Discrete Uniform Distribution

If a random variable X can take on k different values with equal probability, we say that X has a discrete uniform distribution, which is defined as follows:

Definition 4.28

Let X be a discrete random variable taking the values 1, 2, 3, …., n.

X is said to follow a uniform distribution if its pmf is given by

$\begin{array}{l} P (X = x) & = \frac{1}{k}, & x = 1, 2, 3, \dots k \\ = 0, & otherwise \end{array}$ $\begin{array}{l} P (X = x) & = \frac{1}{k}, & x = 1, 2, 3, \dots k \\ = 0, & otherwise \end{array}$

k is called the parameter of the distribution. Discrete uniform distribution is applied whenever all the values of the random variable X are equally likely. The mean of the discrete uniform distribution is $\frac{k + 1}{2}$ $\frac{k + 1}{2}$ and the variance is $\frac{k^{2} - 1}{12}$ $\frac{k^{2} - 1}{12}$ . Coefficient of skewness of the uniform distribution is zero. Hence uniform distribution is symmetric.

Example 4.77: Let X denote the number on the face of a n unbiased die, when it is rolled.

We have

$\begin{array}{l} P (X = x) & = \frac{1}{6}, & x = 1, 2, 3, \dots 6 \\ = 0, & otherwise \end{array}$ $\begin{array}{l} P (X = x) & = \frac{1}{6}, & x = 1, 2, 3, \dots 6 \\ = 0, & otherwise \end{array}$

Clearly X follows uniform distribution.

4.23 The Negative Binomial and Geometric Distribution

Consider an experiment consisting of repeated Bernoulli (independent) trials. The number of the trial on which kth success occurs is a random variable having a negative binomial distribution which defined as follows:

Definition 4.29

A random variable X has a negative binomial distribution if and only if

$P (X = x) = {}^{x - 1}C_{k - 1} p^{k} q^{x - k}, for x = k, k + 1, k + 2, \dots$ $P (X = x) = {}^{x - 1}C_{k - 1} p^{k} q^{x - k}, for x = k, k + 1, k + 2, \dots$

where p is the probability of successes, q in the probability of failure given by q=1−p.

In this case the random variable X is called the negative binomial random variable.

Negative binomial distribution is also called as Pascal or waiting time distribution.

The mean of negative binomial distribution is $k q / p$ $k q / p$ and the variance of negative binomial distributions $k (1 - p) / p^{2}$ $k (1 - p) / p^{2}$ .

The mgf of a negative binomial distribution is $p^{k} {(1 - q e^{t})}^{- k}$ $p^{k} {(1 - q e^{t})}^{- k}$ for a negative binomial distribution Var [X]>E(X). If k=1, the distribution is called geometrical distribution.

4.24 Geometric Distribution

Definition 4.30

A random variable X is said to follow geometric distribution, if it assumes only nonnegative vales and its pmf is

$P (X = x) = p q^{x - 1}, x = 1, 2, 3, \dots where q = 1 - p$ $P (X = x) = p q^{x - 1}, x = 1, 2, 3, \dots where q = 1 - p$

The mean of geometric distribution is $1 / p$ $1 / p$ and the variance is $2 q / p^{2}$ $2 q / p^{2}$ .

The mgf of geometric distribution is $p e^{t} / (1 - (1 - p) e^{t})$ $p e^{t} / (1 - (1 - p) e^{t})$ .

Example 4.78: A typist types 3 letters erroneously for every 100 letters. What is the probability that 10th letter typed is the first erroneous letter?

Solution: We have

$\begin{array}{l} P & = \frac{3}{100} = 0.03 \\ q & = 1 - p = 0.97 \\ x & = 10 \\ P (X = x) & = p q^{x - 1} \\ = (0.03) {(0.97)}^{10 - 1} \\ = (0.03) {(0.97)}^{9} \end{array}$ $\begin{array}{l} P & = \frac{3}{100} = 0.03 \\ q & = 1 - p = 0.97 \\ x & = 10 \\ P (X = x) & = p q^{x - 1} \\ = (0.03) {(0.97)}^{10 - 1} \\ = (0.03) {(0.97)}^{9} \end{array}$

Exercise 4.9

1. If the probability is 0.75 that an applicant for a driver’s license will pass the road test an any given try, what is the probability that an applicant will finally pass the test on fifth try?

Ans: (0.75)(0.25)⁴

2. A typist types 2 letters erroneously for every 100 letters, what is the probability that the truth letter typed is the first erroneously letter?

Ans: (0.02)(0.98)⁹

3. If the probability is 0.40, that a child exposed to a certain contagious disease will catch it, what is the probability that the 10th child exposed to the disease will be the third to catch it?

Ans: ${}^{9}C_{2} {(0.40)}^{3} {(0.60)}^{7}$ ${}^{9}C_{2} {(0.40)}^{3} {(0.60)}^{7}$

4. Find the probability that in tossing four coins one will get either all heads or tails for the third time on seventh toss?

Ans: ${}^{6}C_{2} {(\frac{1}{8})}^{3} {(\frac{7}{8})}^{3}$ ${}^{6}C_{2} {(\frac{1}{8})}^{3} {(\frac{7}{8})}^{3}$

5. The probability of a student passing a subject is 0.6. What is the probability that he will pass in the subject in his third attempt?

Ans: (0.6)(0.4)²

6. In a company, a group of particular items is accepted if more items are tested before the first defective is found. If m=5, and the group consists of 15% defective items, what is the probability that it will be accepted?

Ans: 0.04437

4.25 Continuous Probability Distributions

In this section we discuss the method of finding the probability distributions associated with continuous random variables and also how to use the mean and SD to describe these distributions.

4.25.1 Uniform Distribution

Let X be a continuous random variable. X is said to have uniform distribution, if its pdf is given by

$\begin{array}{l} f (x) & = k, & a < x < b \\ = 0, & otherwise \end{array}$ $\begin{array}{l} f (x) & = k, & a < x < b \\ = 0, & otherwise \end{array}$

where k is a constant given by $\int_{a}^{b} f (x) d x = 1$ $\int_{a}^{b} f (x) d x = 1$

i.e.,

$\int_{a}^{b} k d x = 1$ $\int_{a}^{b} k d x = 1$

$k {[x]}_{a}^{b} = 1$ $k {[x]}_{a}^{b} = 1$

$k (b - a) = 1$ $k (b - a) = 1$

$k = \frac{1}{b - a}$ $k = \frac{1}{b - a}$

Hence

$\begin{array}{l} f (x) & = \frac{1}{b - a}, & a < x < b \\ = 0, & otherwise \end{array}$ $\begin{array}{l} f (x) & = \frac{1}{b - a}, & a < x < b \\ = 0, & otherwise \end{array}$

Uniform distribution is also known as rectangular distribution.

The distribution function f(x) of X is given by

$F (x) = {\begin{array}{l} 0, & - \infty < x < a \\ \frac{x - a}{b - a}, & a \leq x \leq b \\ 1, & b < x < \infty \end{array}$ $F (x) = {\begin{array}{l} 0, & - \infty < x < a \\ \frac{x - a}{b - a}, & a \leq x \leq b \\ 1, & b < x < \infty \end{array}$

Clearly F(x) is not continuous at the two end points, x=a and x=b. Hence F(x) is not differentiable at x=a and x=b.

The pdf of a uniform variate X is (−a, a) is given by

$f (x) = {\begin{array}{l} \frac{1}{2 a}, & - a < x < a \\ 0, & otherwise \end{array}$ $f (x) = {\begin{array}{l} \frac{1}{2 a}, & - a < x < a \\ 0, & otherwise \end{array}$

4.25.1.1 Moments of the Uniform Distribution

We have ${μ'}_{r} = r th$ ${μ'}_{r} = r th$ moment about the origin

$\begin{array}{l} = \int_{a}^{b} x^{r} f (x) d x \\ = \int_{a}^{b} x^{r} \frac{1}{b - a} d x \\ = \frac{1}{b - a} {[\frac{x^{r + 1}}{r + 1}]}_{a}^{b} \\ = \frac{b^{r + 1} - a^{r + 1}}{(r + 1) (b - a)} \end{array}$ $\begin{array}{l} = \int_{a}^{b} x^{r} f (x) d x \\ = \int_{a}^{b} x^{r} \frac{1}{b - a} d x \\ = \frac{1}{b - a} {[\frac{x^{r + 1}}{r + 1}]}_{a}^{b} \\ = \frac{b^{r + 1} - a^{r + 1}}{(r + 1) (b - a)} \end{array}$

Putting r=1, 2, 3, 4 we obtain,

$\begin{array}{l} {μ'}_{1} & = \frac{b^{2} - a^{2}}{2 (b - a)} = \frac{b + a}{2} \\ {μ'}_{2} & = \frac{b^{3} - a^{3}}{3 (b - a)} = \frac{b^{2} + b a + a^{2}}{2} \\ {μ'}_{3} & = \frac{b^{4} - a^{4}}{4 (b - a)} = \frac{(b^{2} + a^{2}) (b + a)}{4} \\ {μ'}_{4} & = \frac{b^{5} - a^{5}}{5 (b - a)} = \frac{b^{4} + b^{3} a + b^{2} a^{2} + b a^{3} + a^{4}}{5} \end{array}$ $\begin{array}{l} {μ'}_{1} & = \frac{b^{2} - a^{2}}{2 (b - a)} = \frac{b + a}{2} \\ {μ'}_{2} & = \frac{b^{3} - a^{3}}{3 (b - a)} = \frac{b^{2} + b a + a^{2}}{2} \\ {μ'}_{3} & = \frac{b^{4} - a^{4}}{4 (b - a)} = \frac{(b^{2} + a^{2}) (b + a)}{4} \\ {μ'}_{4} & = \frac{b^{5} - a^{5}}{5 (b - a)} = \frac{b^{4} + b^{3} a + b^{2} a^{2} + b a^{3} + a^{4}}{5} \end{array}$

4.25.1.2 Mean of Uniform Distribution

${μ'}_{1} = E (X) = \frac{b + a}{2}$ ${μ'}_{1} = E (X) = \frac{b + a}{2}$

4.25.1.3 Variance of Uniform Distribution

$\begin{array}{l} μ_{2} & = {μ'}_{2} - μ_{1}^{' 2} \frac{b^{2} + a b + a^{2}}{3} - \frac{{(b + a)}^{2}}{4} \\ = \frac{1}{12} [4 b^{2} + 4 a b + 4 a^{2} - 3 b^{2} - 3 a^{2} - 6 a b] \\ = \frac{1}{12} [b^{2} + a^{2} - 2 a b] \\ = \frac{{(b - a)}^{2}}{12} \end{array}$ $\begin{array}{l} μ_{2} & = {μ'}_{2} - μ_{1}^{' 2} \frac{b^{2} + a b + a^{2}}{3} - \frac{{(b + a)}^{2}}{4} \\ = \frac{1}{12} [4 b^{2} + 4 a b + 4 a^{2} - 3 b^{2} - 3 a^{2} - 6 a b] \\ = \frac{1}{12} [b^{2} + a^{2} - 2 a b] \\ = \frac{{(b - a)}^{2}}{12} \end{array}$

4.25.1.4 Moment Generating Function of the Uniform Distribution

We have

$\begin{array}{l} M (t) & = E [e^{t x}] \\ = \int_{a}^{b} e^{t x} \frac{1}{b - a} d x = \frac{e^{b t} e^{a t}}{(b - a) t} \\ = \frac{(1 + \frac{b t}{1!} + \frac{b^{2} t^{2}}{2!} + \dots) - (1 + \frac{a t}{1!} + \frac{a^{2} t^{2}}{2!} + \dots)}{(b - a) t} \\ = 1 + \frac{b + a}{2} \cdot t + \frac{b^{3} - a^{3}}{3 (b - a)} \cdot \frac{t^{2}}{2} + \frac{b^{4} - a^{4}}{4 (b - a)} \cdot \frac{t^{3}}{3} + \dots \end{array}$ $\begin{array}{l} M (t) & = E [e^{t x}] \\ = \int_{a}^{b} e^{t x} \frac{1}{b - a} d x = \frac{e^{b t} e^{a t}}{(b - a) t} \\ = \frac{(1 + \frac{b t}{1!} + \frac{b^{2} t^{2}}{2!} + \dots) - (1 + \frac{a t}{1!} + \frac{a^{2} t^{2}}{2!} + \dots)}{(b - a) t} \\ = 1 + \frac{b + a}{2} \cdot t + \frac{b^{3} - a^{3}}{3 (b - a)} \cdot \frac{t^{2}}{2} + \frac{b^{4} - a^{4}}{4 (b - a)} \cdot \frac{t^{3}}{3} + \dots \end{array}$

Example 4.79: If X is uniformly distributed over $(- a, a)$ $(- a, a)$ . Find a so that $P (X > 1) = 1 / 3$ $P (X > 1) = 1 / 3$ .

Solution: Therefore the pdf of X is

$f (x) = {\begin{array}{l} \frac{1}{30}, & 0 < x < 30 \\ 0, & otherwise \end{array}$ $f (x) = {\begin{array}{l} \frac{1}{30}, & 0 < x < 30 \\ 0, & otherwise \end{array}$

We have

$P (X > 1) = \frac{1}{3} (given)$ $P (X > 1) = \frac{1}{3} (given)$

i.e.,

$\int_{1}^{a} f (x) d x = \frac{1}{3}$ $\int_{1}^{a} f (x) d x = \frac{1}{3}$

$\int_{1}^{a} \frac{1}{2 a} d x = \frac{1}{3}$ $\int_{1}^{a} \frac{1}{2 a} d x = \frac{1}{3}$

$\frac{1}{2 a} {[x]}_{1}^{a} = \frac{1}{3}$ $\frac{1}{2 a} {[x]}_{1}^{a} = \frac{1}{3}$

$\frac{a - 1}{2 a} = \frac{1}{3}$ $\frac{a - 1}{2 a} = \frac{1}{3}$

i.e., 3a–3=2a

Hence a=3

Example 4.80: Subway trains on a certain line run after every half an hour between midnight and 6:00 in the morning. What is the probability that a man entering the station at a random time during this period will have to wait at least 20 minutes?

Solution: Let the waiting time for the next train by the man be denoted by X.

Clearly X is a random variable which in uniformly distributed in (0, 30) when the man arrived at station. The pdf of X is given by

$f (x) = {\begin{array}{l} \frac{1}{30}, & 0 < x < 30 \\ 0, & otherwise \end{array}$ $f (x) = {\begin{array}{l} \frac{1}{30}, & 0 < x < 30 \\ 0, & otherwise \end{array}$

Probability that a man entering the station at random waits at least 20 minutes

$\begin{array}{l} = P (X \geq 20) \\ = \int_{20}^{30} f (x) d x = \int_{20}^{30} \frac{1}{30} d x = \frac{1}{30} {[x]}_{20}^{30} = \frac{30 - 20}{30} = \frac{1}{3} \end{array}$ $\begin{array}{l} = P (X \geq 20) \\ = \int_{20}^{30} f (x) d x = \int_{20}^{30} \frac{1}{30} d x = \frac{1}{30} {[x]}_{20}^{30} = \frac{30 - 20}{30} = \frac{1}{3} \end{array}$

4.25.2 Exponential and Negative Exponential Distribution

Let X be a continuous random variable, X is said to have an exponential distribution if the pdf of x is given by

$f (x) = {\begin{array}{l} e^{x}, & x \geq 0 \\ 0, & otherwise \end{array}$ $f (x) = {\begin{array}{l} e^{x}, & x \geq 0 \\ 0, & otherwise \end{array}$

and X is said to have a negative exponential distribution with parameter $λ > 0$ $λ > 0$ . If the pdf is given by

$f (x) = {\begin{array}{l} λ e^{- λ x}, & x > 0 \\ 0, & x \leq 0 \end{array}$ $f (x) = {\begin{array}{l} λ e^{- λ x}, & x > 0 \\ 0, & x \leq 0 \end{array}$

Mean of negative exponential distribution is $1 / λ$ $1 / λ$ and the variance is $1 / λ^{2}$ $1 / λ^{2}$ , the mgf of the negative exponential distribution is $\frac{λ}{λ - t}, (λ > t)$ $\frac{λ}{λ - t}, (λ > t)$ rth moment about origin is $r! / λ^{r}$ $r! / λ^{r}$ , r=1, 2, 3,….

4.26 Normal Distribution

The Normal distribution is a continuous distribution. It was first discovered by Abraham De moivre (1667–1754). He derived normal probability function as the limiting form of the binomial distribution, simultaneously Karl Friedrich Gauss (1777–1853) has also derived normal distribution as law of errors. Hence the normal probability distribution is also called as the Gaussian distribution.

Definition 4.31

A continuous random variable X is said to have a Normal distribution if its probability function is given by

$f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}, (- \infty < x < \infty)$ $f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}, (- \infty < x < \infty)$

where μ=mean of the normal random variable; σ=Standard deviation of the normal variable X; σ²=Variance of the normal variable X and are the parameters of the distribution. If X follows Normal distribution then it is denoted by $X ~ N (μ, σ)$ $X ~ N (μ, σ)$ .

4.26.1 Standard Normal Variable

Computing the area over intervals under the normal probability distribution is difficult task. Consequently we will use a table of areas as a function of Z-score. The Z-score give the distance between the measurement and the mean in units equal to the SD. Z is always a normal random variable with mean and SD 1. For this reason Z is often referred to as the standard normal random variable. Z is a continuous random variable.

Definition 4.32

The standard normal random variable Z is defined by

$Z = \frac{x - μ}{σ}$ $Z = \frac{x - μ}{σ}$

where μ=mean of the normal random variable X; σ=Standard deviation of the normal variable X; X=normal random variable.

Definition 4.33

A random variable $Z = (x - μ) / σ$ $Z = (x - μ) / σ$ is said to have a standard normal distribution if its probability function is defined by

$f (z) = \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} Z^{2}}, (- \infty < Z < \infty)$ $f (z) = \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} Z^{2}}, (- \infty < Z < \infty)$

A standard normal variate is denoted by N(0, 1). Standard normal distribution is also known as Z-distribution or unit Normal distribution. Standardization of Normal distribution helps us to make use of the tables of area of standard curve.

$\frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} Z^{2}}$ $\frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} Z^{2}}$

For various points along the x-axis

The area between the points say Z₁ and Z₂ under the standard normal curve will represent the probability, the z will lie between Z₁ and Z₂. It is denoted by

$P (Z_{1} \leq Z \leq Z_{2})$ $P (Z_{1} \leq Z \leq Z_{2})$

The general equation of the normal curve is

$y = f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}, (- \infty < x < \infty)$ $y = f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}, (- \infty < x < \infty)$

If the corresponding total frequency is N then

$y = f (x) = \frac{N}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}$ $y = f (x) = \frac{N}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}$

is the equation if best fitting normal probability curve.

4.26.2 Distribution Function $φ (Z)$ $φ (Z)$ of Standard Normal Variate

The distribution function $φ (Z)$ $φ (Z)$ of standard normal variate, Z is defined by properties of

$\begin{array}{l} φ (Z) & = P (Z \leq z) = \int_{- \infty}^{Z} f (t) d t \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{Z} \cdot e^{- \frac{1}{2} Z^{2}} \end{array}$ $\begin{array}{l} φ (Z) & = P (Z \leq z) = \int_{- \infty}^{Z} f (t) d t \\ = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{Z} \cdot e^{- \frac{1}{2} Z^{2}} \end{array}$

where

$Z = \frac{x - μ}{σ}$ $Z = \frac{x - μ}{σ}$

Properties of $φ (Z)$ $φ (Z)$ :

1. $φ (- Z) = 1 - φ (Z)$ $φ (- Z) = 1 - φ (Z)$

2. $P (a \leq X \leq b) = φ (\frac{b - μ}{σ}) - φ (\frac{a - μ}{σ})$ $P (a \leq X \leq b) = φ (\frac{b - μ}{σ}) - φ (\frac{a - μ}{σ})$

3. P(Z≤a)=P(Z≥−a)

4.26.3 Area Under Normal Curve

The curve of normal distribution is unimodal and is bell shaped with the highest point over the mean μ. It is symmetrical about a vertical line through μ.

The normal curve has the following features (Fig. 4.2):

1. The curve is symmetrical about the coordinate at its mean, which locates the peak of the bell.

2. The values of mean, median, and mode are equal.

3. The curve extends from $- \infty$ $- \infty$ to $\infty$ $\infty$ .

4. The area covered between $μ - σ$ $μ - σ$ and $μ + σ$ $μ + σ$ is 0.6826 (i.e., 68.26% of the total area).

5. The area covered between the limits $μ - 2 σ$ $μ - 2 σ$ and $μ + 2 σ$ $μ + 2 σ$ is 0.9544 (i.e., 95.44% of area).

6. The area covered between $μ - 3 σ$ $μ - 3 σ$ and $μ + 3 σ$ $μ + 3 σ$ is 0.9974 (i.e., 99.74% of area).

4.26.4 Area Under Standard Normal Curve

Let X be a normal random variable and if Z be the corresponding standard normal variable then Z and X have identical curves. The curve of Z is called standard normal curve. The area bounded by a standard normal variable Z, the Z axis and the ordinates at $Z = \infty$ $Z = \infty$ and any positive value of Z is provided by a standard table.

The standard normal curve covers 68.26% area between Z=−1 and 1.

Therefore

$P (- 1 \leq Z \leq 1) = 0.6826$ $P (- 1 \leq Z \leq 1) = 0.6826$

The curve covers 95.44% of the total area between Z=−2 and 2.

Therefore

$P (- 2 \leq Z \leq 2) = 0.9544$ $P (- 2 \leq Z \leq 2) = 0.9544$

And the area covered between Z=−3 and 3 is 99.74% of the total area.

Hence

$P (- 3 \leq Z \leq 3) = 0.9474$ $P (- 3 \leq Z \leq 3) = 0.9474$

4.26.5 Properties of Normal Curve

1. The mean μ and variance σ² of a normal distribution are called the parameters of the distribution.

2. The mean, median, and mode of a normal distribution coincide with each other.

3. In a normal distribution, mean deviation about mean is approximately equal to $4 / 3$ $4 / 3$ times its SD.

4. In a normal distribution, the quartiles Q₁ and Q₃ are equidistant from the median.

5. The normal curve is bell shaped and is symmetrical about $X = μ$ $X = μ$ .

6. The area bound by the normal curve and x-axis equal to 1.

7. The tails of the curve of a normal distribution extend identically in both sides of $x = μ$ $x = μ$ and never touch the x-axis.

8. The probability that x lies between a and b is equal to the area bounded by the curve X-axis and ordinates X=a and X=b.

4.26.6 Mean of Normal Distribution

Consider the normal distribution with as the parameter then,

Mean of X=E[x]

Therefore

$\begin{array}{l} μ & = E [X] = \int_{- \infty}^{\infty} x f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$ $\begin{array}{l} μ & = E [X] = \int_{- \infty}^{\infty} x f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$

Putting

$Z = \frac{x - μ}{σ}$ $Z = \frac{x - μ}{σ}$ , we have

$d Z = \frac{d x}{σ}$ $d Z = \frac{d x}{σ}$ i.e., $d x = σ d Z$ $d x = σ d Z$

and

$x = μ + σ Z$ $x = μ + σ Z$

Hence we get

$\begin{array}{l} μ & = \frac{1}{\sqrt{2 π}} {\int_{- \infty}^{\infty} (σ z + μ) e}^{- \frac{1}{2} Z^{2}} d z \\ = \frac{σ}{\sqrt{2 π}} {\int_{\infty}^{\infty} Z e}^{- \frac{1}{2} z^{2}} d z + \frac{μ}{\sqrt{2 π}} {\int_{- \infty}^{\infty} e}^{- \frac{1}{2} z^{2}} d z \\ = \frac{σ}{\sqrt{2 π}} {\int_{- \infty}^{\infty} Z e}^{- \frac{1}{2} z^{2}} d z + \frac{2 μ}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} d z \\ = \frac{\sqrt{2} μ}{\sqrt{π}} \cdot \frac{\sqrt{π}}{\sqrt{2}} = μ \end{array}$ $\begin{array}{l} μ & = \frac{1}{\sqrt{2 π}} {\int_{- \infty}^{\infty} (σ z + μ) e}^{- \frac{1}{2} Z^{2}} d z \\ = \frac{σ}{\sqrt{2 π}} {\int_{\infty}^{\infty} Z e}^{- \frac{1}{2} z^{2}} d z + \frac{μ}{\sqrt{2 π}} {\int_{- \infty}^{\infty} e}^{- \frac{1}{2} z^{2}} d z \\ = \frac{σ}{\sqrt{2 π}} {\int_{- \infty}^{\infty} Z e}^{- \frac{1}{2} z^{2}} d z + \frac{2 μ}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} d z \\ = \frac{\sqrt{2} μ}{\sqrt{π}} \cdot \frac{\sqrt{π}}{\sqrt{2}} = μ \end{array}$

4.26.7 Variance of Normal Distribution

$Variance = E {(X - μ)}^{2}$ $Variance = E {(X - μ)}^{2}$

i.e.,

$\begin{array}{l} V [X] & = \int_{- \infty}^{\infty} {(x - μ)}^{2} f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} {(x - μ)}^{2} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$ $\begin{array}{l} V [X] & = \int_{- \infty}^{\infty} {(x - μ)}^{2} f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} {(x - μ)}^{2} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$

Putting $Z = \frac{x - μ}{σ}$ $Z = \frac{x - μ}{σ}$ , we get

$V [X] = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} σ^{2} z^{2} e^{- \frac{1}{2} z^{2}} (σ d z)$ $V [X] = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} σ^{2} z^{2} e^{- \frac{1}{2} z^{2}} (σ d z)$

Putting $Z^{2} / 2 = t$ $Z^{2} / 2 = t$ , we obtain

$\begin{array}{l} V [X] & = \frac{2 σ^{2}}{\sqrt{2 π}} \int_{0}^{\infty} 2 t e^{- t} \frac{d t}{\sqrt{2 t}} = \frac{2 σ^{2}}{\sqrt{π}} \int_{0}^{\infty} e^{- t} \sqrt{t} d t \\ = \frac{2 σ^{2}}{\sqrt{π}} \int_{0}^{\infty} e^{- t} t^{\frac{3}{2} - 1} d t \\ = \frac{2 σ^{2}}{\sqrt{π}} Γ (\frac{3}{2}) = \frac{2 σ^{2}}{\sqrt{π}} \frac{3}{2} Γ (\frac{1}{2}) \\ = \frac{σ^{2}}{\sqrt{π}} . \sqrt{π} \end{array}$ $\begin{array}{l} V [X] & = \frac{2 σ^{2}}{\sqrt{2 π}} \int_{0}^{\infty} 2 t e^{- t} \frac{d t}{\sqrt{2 t}} = \frac{2 σ^{2}}{\sqrt{π}} \int_{0}^{\infty} e^{- t} \sqrt{t} d t \\ = \frac{2 σ^{2}}{\sqrt{π}} \int_{0}^{\infty} e^{- t} t^{\frac{3}{2} - 1} d t \\ = \frac{2 σ^{2}}{\sqrt{π}} Γ (\frac{3}{2}) = \frac{2 σ^{2}}{\sqrt{π}} \frac{3}{2} Γ (\frac{1}{2}) \\ = \frac{σ^{2}}{\sqrt{π}} . \sqrt{π} \end{array}$

Hence variance=σ²

4.26.8 Mode of Normal Distribution

We have

$f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}$ $f (x) = \frac{1}{σ \sqrt{2 π}} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}}$

Applying logarithms on both sides, we get

$Log f (x) = - \log σ \sqrt{2 π} - \frac{1}{2 σ^{2}} \cdot {(x - μ)}^{2}$ $Log f (x) = - \log σ \sqrt{2 π} - \frac{1}{2 σ^{2}} \cdot {(x - μ)}^{2}$

Differentiating with respect to x we get

$\frac{f' (x)}{f (x)} = - \frac{1}{σ^{2}} (x - μ)$ $\frac{f' (x)}{f (x)} = - \frac{1}{σ^{2}} (x - μ)$

$f' (x) = - \frac{1}{σ^{2}} (x - μ) f (x)$ $f' (x) = - \frac{1}{σ^{2}} (x - μ) f (x)$ (4.17)

(4.17)

Again differentiating, we get

$\begin{array}{l} f ″ (x) & = - \frac{1}{σ^{2}} [1 . f (x) + (x - μ) f' (x)] \end{array}$ $\begin{array}{l} f ″ (x) & = - \frac{1}{σ^{2}} [1 . f (x) + (x - μ) f' (x)] \end{array}$

$= \frac{f (x)}{σ^{2}} [1 - {(\frac{(x - μ)}{σ})}^{2}]$ $= \frac{f (x)}{σ^{2}} [1 - {(\frac{(x - μ)}{σ})}^{2}]$ using Eq. 4.17

f ′(x)=0 gives x=μ (since f(x) ≠ 0)

At x=μ, we have

$\begin{array}{l} f ″ (x) & = \frac{1}{σ^{2}} f (u) \\ = - \frac{1}{σ^{2}} \frac{1}{σ \sqrt{2 π}} \\ = - \frac{1}{σ^{3} \sqrt{2 π}} < 0 \end{array}$ $\begin{array}{l} f ″ (x) & = \frac{1}{σ^{2}} f (u) \\ = - \frac{1}{σ^{2}} \frac{1}{σ \sqrt{2 π}} \\ = - \frac{1}{σ^{3} \sqrt{2 π}} < 0 \end{array}$

Hence x=μ is the mode of the distribution.

4.26.9 Median of the Normal Distribution

Let M denote the median of normal distribution

$\int_{- \infty}^{M} f (x) d x = \frac{1}{2}$ $\int_{- \infty}^{M} f (x) d x = \frac{1}{2}$

i.e.,

$\frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x + \frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = \frac{1}{2}$ $\frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x + \frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = \frac{1}{2}$ (4.18)

(4.18)

Consider

$\frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x$ $\frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x$

Putting

$\frac{x - μ}{σ} = Z$ $\frac{x - μ}{σ} = Z$ , we get dx=σdz

Therefore

$\begin{array}{l} \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x & = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{0} e^{- \frac{1}{2} z^{2}} d z \\ = \frac{1}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} d z (by symmetry) \\ = \frac{1}{\sqrt{2 π}} \frac{\sqrt{π}}{\sqrt{2}} = \frac{1}{2} \end{array}$ $\begin{array}{l} \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{μ} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x & = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{0} e^{- \frac{1}{2} z^{2}} d z \\ = \frac{1}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} d z (by symmetry) \\ = \frac{1}{\sqrt{2 π}} \frac{\sqrt{π}}{\sqrt{2}} = \frac{1}{2} \end{array}$ (4.19)

(4.19)

from Eqs. (4.18) and (4.19), we obtain

$\frac{1}{2} + \frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = \frac{1}{2}$ $\frac{1}{2} + \frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = \frac{1}{2}$

thus

$\frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = 0$ $\frac{1}{σ \sqrt{2 π}} \int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = 0$

$\int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = 0$ $\int_{μ}^{M} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x = 0$

which is possible when μ=M

Hence

$μ = M$ $μ = M$

i.e., Median of normal distribution=μ

4.26.10 Moment Generating Function of Normal Distribution With Respect to Origin

We have

$\begin{array}{l} M_{x} (t) & = E (e^{t x}) = \int_{- \infty}^{\infty} e^{t x} f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{t x} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$ $\begin{array}{l} M_{x} (t) & = E (e^{t x}) = \int_{- \infty}^{\infty} e^{t x} f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{t x} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} d x \end{array}$

Putting

$\frac{x - μ}{σ} = Z$ $\frac{x - μ}{σ} = Z$ , we get

$\begin{array}{l} M_{x} (t) & = \frac{e^{μ t}}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} z^{2} + {(t σ)}^{2}} d z \\ = \frac{e^{μ t + \frac{1}{2} t^{2} σ^{2}}}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {(z^{2} - t σ)}^{2}} d z \\ = \frac{e^{μ t + \frac{1}{2} t^{2} σ^{2}}}{\sqrt{2 π}} \cdot 2 \frac{\sqrt{π}}{\sqrt{2}} \\ = e^{μ t + \frac{1}{2} t^{2} σ^{2}} \end{array}$ $\begin{array}{l} M_{x} (t) & = \frac{e^{μ t}}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} z^{2} + {(t σ)}^{2}} d z \\ = \frac{e^{μ t + \frac{1}{2} t^{2} σ^{2}}}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {(z^{2} - t σ)}^{2}} d z \\ = \frac{e^{μ t + \frac{1}{2} t^{2} σ^{2}}}{\sqrt{2 π}} \cdot 2 \frac{\sqrt{π}}{\sqrt{2}} \\ = e^{μ t + \frac{1}{2} t^{2} σ^{2}} \end{array}$

4.26.11 Mean Deviation of Normal Distribution

$\begin{array}{l} We have mean deviation & = \int_{- \infty}^{\infty} | x - μ | f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} | x - μ | d x \\ = \frac{σ}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} z^{2}} | z | d z where Z = \frac{x - μ}{σ}, \\ = \frac{2 σ}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} | z | d z (since the integral is even) \\ = σ \sqrt{\frac{2}{π} = \frac{4}{5} σ} (approximately) \end{array}$ $\begin{array}{l} We have mean deviation & = \int_{- \infty}^{\infty} | x - μ | f (x) d x \\ = \frac{1}{σ \sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} {[\frac{x - μ}{σ}]}^{2}} | x - μ | d x \\ = \frac{σ}{\sqrt{2 π}} \int_{- \infty}^{\infty} e^{- \frac{1}{2} z^{2}} | z | d z where Z = \frac{x - μ}{σ}, \\ = \frac{2 σ}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{1}{2} z^{2}} | z | d z (since the integral is even) \\ = σ \sqrt{\frac{2}{π} = \frac{4}{5} σ} (approximately) \end{array}$

4.26.11.1 Solved Examples

Example 4.81: A normal distribution has a mean 20 and a SD of 4. Find out the probability that a value of X lies between 20 and 24?

Solution: We have μ=20, and σ=4 (given)

$Z = \frac{x - μ}{σ} = \frac{x - 20}{4}$ $Z = \frac{x - μ}{σ} = \frac{x - 20}{4}$

When

$X = 20, Z = \frac{20 - 20}{4} = 0$ $X = 20, Z = \frac{20 - 20}{4} = 0$

And when

$X = 24, Z = \frac{24 - 20}{4} = 1$ $X = 24, Z = \frac{24 - 20}{4} = 1$

Therefore Z lies between 20 and 24, the value of Z lies between 0 and 1.

Hence

$P (0 \leq Z \leq 1) = 0.3413$ $P (0 \leq Z \leq 1) = 0.3413$

Example 4.82: The mean and SD of a normal variable x are 50 and 4, respectively. Find the values of the corresponding standard normal variable when x is equal to 42, 54, and 84?

Solution: We have

μ=50, and σ=4 (given)

$Z = \frac{x - μ}{σ} = \frac{x - 50}{4}$ $Z = \frac{x - μ}{σ} = \frac{x - 50}{4}$

when

$x = 42, Z = \frac{42 - 50}{4} = - 2$ $x = 42, Z = \frac{42 - 50}{4} = - 2$

when

$x = 54, Z = \frac{54 - 50}{4} = 1$ $x = 54, Z = \frac{54 - 50}{4} = 1$

and when

$x = 84, Z = \frac{84 - 50}{4} = 8.5$ $x = 84, Z = \frac{84 - 50}{4} = 8.5$

Example 4.83: Find the area under the standard normal curve which lies

1. to the left of Z=1.73.

2. to the right of Z=−0.66.

3. between Z=1.25 and Z=1.67.

4. between Z=−1.45 and Z=1.45.

Solution:

1. Area to the left of Z=1.73

$\begin{array}{l} = 0.5 + area between Z = 0 and Z = 1.73 \\ = 0.5 + 0.4582 \\ = 0.9582 (using the table) \end{array}$ $\begin{array}{l} = 0.5 + area between Z = 0 and Z = 1.73 \\ = 0.5 + 0.4582 \\ = 0.9582 (using the table) \end{array}$

2. Area to the right of Z=−0.66

$\begin{array}{l} = area from Z = - 0.66 to Z = 0 + 0.5 \\ = 0.2454 + 0.5 \\ = 0.7454 (using the table) \end{array}$ $\begin{array}{l} = area from Z = - 0.66 to Z = 0 + 0.5 \\ = 0.2454 + 0.5 \\ = 0.7454 (using the table) \end{array}$

3. Area between Z=1.25 and 1.67

$\begin{array}{l} = Area between Z = 0 and 1.67 \\ - (Area between Z = 0 and 1.25) \\ = 0.4525 - 0.3944 \\ = 0.0581 (using the table) \end{array}$ $\begin{array}{l} = Area between Z = 0 and 1.67 \\ - (Area between Z = 0 and 1.25) \\ = 0.4525 - 0.3944 \\ = 0.0581 (using the table) \end{array}$

4. Area between Z=−1.45 and 1.45

$\begin{array}{l} = Area between Z = 0 and 1.45 \\ - (Area between Z = 0 and 1.45) (by symmetry) \\ = 2 (Area between Z = 0 and 1.45) \\ = 2 (0.4265) \\ = 0.8530 (using the table) \end{array}$ $\begin{array}{l} = Area between Z = 0 and 1.45 \\ - (Area between Z = 0 and 1.45) (by symmetry) \\ = 2 (Area between Z = 0 and 1.45) \\ = 2 (0.4265) \\ = 0.8530 (using the table) \end{array}$

Example 4.84: The ABC company uses a machine to fill boxes with soap powder. Assume that the net weight of the boxes of soap powder is normally distributed with mean 15 ounces and SD 8 ounces.

1. What proportion of boxes will have net weight of more than 14 ounces?

2. 25% of boxes will be heavier than a certain net weight ω and 75% of the boxes will be lighter than this net weight. Find ω?

Solution: We have μ=15 ounce, σ=0.8 ounce

Let Z be the standard normal variable

$Z = \frac{x - μ}{σ} = \frac{x - 15}{0.8}$ $Z = \frac{x - μ}{σ} = \frac{x - 15}{0.8}$

1. When x=14,

$Z = \frac{14 - 15}{0.8} = \frac{- 1}{0.8} = - 1.25$ $Z = \frac{14 - 15}{0.8} = \frac{- 1}{0.8} = - 1.25$

when

$\begin{array}{l} X & = 14 the probability that X \geq 14 = P (X \geq 14) \\ = P (Z > - 1.25) \\ = P (- 1.25 \leq Z \leq 0) + P (Z \geq 0) \\ = P (0 \leq Z \leq 1.25) + 0.5 \\ = 0.3944 + 0.5 \\ = 0.8944 \end{array}$ $\begin{array}{l} X & = 14 the probability that X \geq 14 = P (X \geq 14) \\ = P (Z > - 1.25) \\ = P (- 1.25 \leq Z \leq 0) + P (Z \geq 0) \\ = P (0 \leq Z \leq 1.25) + 0.5 \\ = 0.3944 + 0.5 \\ = 0.8944 \end{array}$

89.44% boxes will have their net weight greater than 14.

2. Taking x=ω, we have $Z = \frac{x - μ}{σ} = \frac{ω - 15}{0.8}$ $Z = \frac{x - μ}{σ} = \frac{ω - 15}{0.8}$
Probability of Z being greater than $\frac{ω - 15}{0.8}$ $\frac{ω - 15}{0.8}$ is 0.25.
Therefore $0.25 = 0.5 - P (0 \leq Z \leq \frac{ω - 15}{0.8})$ $0.25 = 0.5 - P (0 \leq Z \leq \frac{ω - 15}{0.8})$
i.e., $P (0 \leq Z \leq \frac{ω - 15}{0.8}) = 0.5 - 0.25 = 0.25$ $P (0 \leq Z \leq \frac{ω - 15}{0.8}) = 0.5 - 0.25 = 0.25$
The value of Z for which the probability is 0.25=0.675
Therefore $\frac{ω - 15}{0.8} = 0.694$ $\frac{ω - 15}{0.8} = 0.694$
or $\begin{array}{l} ω & = 0.675 \times 0.8 + 15 \\ = 15.54 ounces \end{array}$ $\begin{array}{l} ω & = 0.675 \times 0.8 + 15 \\ = 15.54 ounces \end{array}$

Example 4.85: Let X denote the number of successes in test. If X is normally distributed with mean 100 and SD 15. Find the probability that x does not exceed 130?

Solution: We have μ=100 and σ=15

Let

$Z = \frac{x - μ}{σ} = \frac{x - 100}{15}$ $Z = \frac{x - μ}{σ} = \frac{x - 100}{15}$

Probability that x does not exceed 130=P(X≤130).

$\begin{array}{l} = P \\ = P (Z \leq 2) \\ = P (- \infty < Z \leq 0) + P (0 \leq Z \leq 2) \\ = 0.5 + 0.4772 \\ = 0.9772 (from the area table) \end{array}$ $\begin{array}{l} = P \\ = P (Z \leq 2) \\ = P (- \infty < Z \leq 0) + P (0 \leq Z \leq 2) \\ = 0.5 + 0.4772 \\ = 0.9772 (from the area table) \end{array}$

Example 4.86: Assume that mean height of soldiers to be 68.22 in. with a variance of 10.8 in. How many soldiers in a regiment of 1000 would you expect to be over 6 ft. tall?

Solution: We have μ=68.22, σ²=10.8, $σ = \sqrt{10.8} = 3.286$ $σ = \sqrt{10.8} = 3.286$

$X = 6 ft . = 6 \times 12 = 72 in .$ $X = 6 ft . = 6 \times 12 = 72 in .$

when

$X = 72, Z = \frac{x - μ}{σ} = \frac{72 - 68.22}{3.286} = 1.1503$ $X = 72, Z = \frac{x - μ}{σ} = \frac{72 - 68.22}{3.286} = 1.1503$

$\begin{array}{l} P (X > 72) & = P (Z > 1.1503) \\ = 0.5 - P (0 < Z < 1.1503) \\ = 0.5 - 0.3749 \\ = 0.1251 \end{array}$ $\begin{array}{l} P (X > 72) & = P (Z > 1.1503) \\ = 0.5 - P (0 < Z < 1.1503) \\ = 0.5 - 0.3749 \\ = 0.1251 \end{array}$

Expected number of soldiers over 6 ft. tall in regiment of

$\begin{array}{l} 1000 & = 1000 \times 0.1251 \\ = 125.1 \\ = 125 soldiers . \end{array}$ $\begin{array}{l} 1000 & = 1000 \times 0.1251 \\ = 125.1 \\ = 125 soldiers . \end{array}$

Example 4.87: What is the probability that a standard normal variate Z will be lying between 1.25 and 2.75?

Solution:

$\begin{array}{l} P (1.25 < Z < 2.75) & = P (0 < Z < 2.75) - P (0 < Z < 1.25) \\ = Area between Z = 0 and Z = 2.75 \\ - (Area between Z = 0 and Z = 1.25) \\ = 0.4970 - 0.3944 \\ = 0.1026 \end{array}$ $\begin{array}{l} P (1.25 < Z < 2.75) & = P (0 < Z < 2.75) - P (0 < Z < 1.25) \\ = Area between Z = 0 and Z = 2.75 \\ - (Area between Z = 0 and Z = 1.25) \\ = 0.4970 - 0.3944 \\ = 0.1026 \end{array}$

Example 4.88: In a distribution exactly 7% of the items are under 35 and 89% of the items are under 63. What are the values of mean and SD of the distribution?

Solution: Let μ be the mean and σ be the SD of the distribution.

The area lying to the left of the ordinate at x=35 is 7%, i.e., 0.07. The corresponding values of Z is negative.

The area lying to the right of the ordinate at x=63 up to the mean μ is

$= 0.5 - 0.07 = 0.43 .$ $= 0.5 - 0.07 = 0.43 .$

Hence the values of Z corresponding in the area 0.43 is 1.48.

i.e.,

$\frac{35 - μ}{σ} = - 1.48 or μ - 1.48 σ = 35$ $\frac{35 - μ}{σ} = - 1.48 or μ - 1.48 σ = 35$ (4.20)

(4.20)

Similarly the area lying to the left of the ordinate at x=63 up to the mean is 0.39 (i.e., 39%).

The value of Z corresponding to the area 0.39 is 1.23.

i.e.,

$\frac{63 - μ}{σ} = 1.23 or μ + 1.23 σ = 63$ $\frac{63 - μ}{σ} = 1.23 or μ + 1.23 σ = 63$ (4.21)

(4.21)

From Eqs. (4.20) and (4.21), we get

$μ = 50.3; σ = 1.33$ $μ = 50.3; σ = 1.33$

Example 4.89: The income of a group of 10,000 persons was found to be normally distributed with mean equal to Rs. 750, and SD equal to Rs. 50. What was lowest income among the richest 250?

Solution: We have μ=750, σ=50 (given)

Let x′ be the lowest income among the richest 250 people.

Then

$P (x \geq x) = \frac{250}{10, 000} (given)$ $P (x \geq x) = \frac{250}{10, 000} (given)$

i.e.,

$P (\frac{x - μ}{σ} \geq \frac{x' - 750}{50}) = 0.025$ $P (\frac{x - μ}{σ} \geq \frac{x' - 750}{50}) = 0.025$

$P (Z \geq \frac{x' - 750}{50}) = 0.025$ $P (Z \geq \frac{x' - 750}{50}) = 0.025$

We obtain

$(\frac{x' - 750}{50}) = 1.96$ $(\frac{x' - 750}{50}) = 1.96$

$x' - 750 = 50 \times 1.96$ $x' - 750 = 50 \times 1.96$

$x' = 750 + 98 = 848$ $x' = 750 + 98 = 848$

Hence Rs. 848 is the lowest income among the richest 250 people.

Example 4.90: If log_e x is normally distributed with mean 1 and variance 4. Find $P (\frac{1}{2} < X < 2)$ $P (\frac{1}{2} < X < 2)$ given that log_e2=0.693?

Solution: Here we have μ=1, variance=σ²=4 (given)

$∴ σ = \sqrt{4} = 2$ $∴ σ = \sqrt{4} = 2$

Let

$\begin{array}{l} X & = \log_{e} x \\ Z & = \frac{x - μ}{σ} = \frac{x - 1}{2} = \frac{\log_{e} x - 1}{2} \end{array}$ $\begin{array}{l} X & = \log_{e} x \\ Z & = \frac{x - μ}{σ} = \frac{x - 1}{2} = \frac{\log_{e} x - 1}{2} \end{array}$

When

$\begin{array}{l} x & = \frac{1}{2}, \\ Z_{1} & = \frac{\log_{e} \frac{1}{2} - 1}{2} = \frac{\log_{e} 1 - \log_{e} 2 - 1}{2} \\ = \frac{- \log_{e} 2 - 1}{2} = \frac{0.693 - 1}{2} \\ = - 0.8465 \end{array}$ $\begin{array}{l} x & = \frac{1}{2}, \\ Z_{1} & = \frac{\log_{e} \frac{1}{2} - 1}{2} = \frac{\log_{e} 1 - \log_{e} 2 - 1}{2} \\ = \frac{- \log_{e} 2 - 1}{2} = \frac{0.693 - 1}{2} \\ = - 0.8465 \end{array}$

When x=2,

$\begin{array}{l} Z_{2} & = \frac{\log_{e} 2 - 1}{2} = \frac{0.693 - 1}{2} \\ = - 0.1535 \end{array}$ $\begin{array}{l} Z_{2} & = \frac{\log_{e} 2 - 1}{2} = \frac{0.693 - 1}{2} \\ = - 0.1535 \end{array}$

$\begin{array}{l} P (\frac{1}{2} < X < 2) & = P (Z_{1} < Z < Z_{2}) \\ = P (- 0.8465 < Z < - 0.1535) \\ = P (0.1535 > Z > 0.8465) \\ = Area between Z = 0 and Z = 0.8465 \\ - (Area between Z = 0 and Z = 0.1535) \\ = 0.2996 - 0.0596 \\ = 0.24 \end{array}$ $\begin{array}{l} P (\frac{1}{2} < X < 2) & = P (Z_{1} < Z < Z_{2}) \\ = P (- 0.8465 < Z < - 0.1535) \\ = P (0.1535 > Z > 0.8465) \\ = Area between Z = 0 and Z = 0.8465 \\ - (Area between Z = 0 and Z = 0.1535) \\ = 0.2996 - 0.0596 \\ = 0.24 \end{array}$

Example 4.91: In a normal distribution, 31% of the items are under 45 and 8% are over 64. Find the mean and SD of the distribution?

Solution: Let X be the normal variable, μ denote the mean, and σ be the SD of X.

Let

$Z = \frac{x - μ}{σ}$ $Z = \frac{x - μ}{σ}$

It is given that

$P (X > 64) = \frac{31}{100} = 0.31$ $P (X > 64) = \frac{31}{100} = 0.31$

And

$P (X > 64) = \frac{8}{100} = 0.08$ $P (X > 64) = \frac{8}{100} = 0.08$

Since

$P (X < 45) < 0.5$ $P (X < 45) < 0.5$

X lies to the left of X=μ, therefore the value of Z is negative say Z₁.

i.e., when x=45, let $Z = \frac{45 - μ}{σ} = - Z_{1}, (Z_{1} > 0) (say)$ $Z = \frac{45 - μ}{σ} = - Z_{1}, (Z_{1} > 0) (say)$

Similarly Z is positive when X=64

Let

$Z = \frac{64 - μ}{σ} = Z_{2} (Z_{2} > 0)$ $Z = \frac{64 - μ}{σ} = Z_{2} (Z_{2} > 0)$

We have

$P (X < 45) = 0.3$ $P (X < 45) = 0.3$

i.e.,

$P (Z < - Z_{1}) = 0.31$ $P (Z < - Z_{1}) = 0.31$

$P (Z > - Z_{1}) = 0.31$ $P (Z > - Z_{1}) = 0.31$

$0.5 - P (0 \leq Z \leq Z_{1}) = 0.31$ $0.5 - P (0 \leq Z \leq Z_{1}) = 0.31$

i.e.,

$\begin{array}{l} P (0 \leq Z \leq Z_{1}) = 0.19 \\ Z_{1} = 0.5 (from the area table) \end{array}$ $\begin{array}{l} P (0 \leq Z \leq Z_{1}) = 0.19 \\ Z_{1} = 0.5 (from the area table) \end{array}$

Therefore

$\frac{45 - μ}{σ} = - 0.5$ $\frac{45 - μ}{σ} = - 0.5$

$45 - μ = - 0.5 σ$ $45 - μ = - 0.5 σ$ (4.22)

(4.22)

Also we have

$P (X > 64) = 0.08$ $P (X > 64) = 0.08$

i.e.,

$P (Z > Z_{2}) = 0.08$ $P (Z > Z_{2}) = 0.08$

$\begin{array}{l} 0.5 - P (0 \leq Z \leq Z_{2}) = 0.08 \\ P (0 \leq Z \leq Z_{2}) = 0.42 \end{array}$ $\begin{array}{l} 0.5 - P (0 \leq Z \leq Z_{2}) = 0.08 \\ P (0 \leq Z \leq Z_{2}) = 0.42 \end{array}$

Hence Z₂=1.4 (from the area table)

Therefore

$\frac{64 - μ}{σ} = 1.4$ $\frac{64 - μ}{σ} = 1.4$

$64 - μ = 1.4 σ$ $64 - μ = 1.4 σ$ (4.23)

(4.23)

Solving Eqs. (4.22) and (4.23) we get

$μ = 50 and σ = 10$ $μ = 50 and σ = 10$

Example 4.92: The mean inside diameter of a sample of 200 washers produced by a machine is 0.502 cm and that SD is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance is the diameter of 0.496–0.508 cm. Otherwise the washers are considered to be defective washers produced by machine, assuming the diameters are naturally distributed.

Solution: we have μ=0.502 cm and σ=0.005 cm

Let z be the standard normal variable

$Z = \frac{x - μ}{σ} = \frac{x - 0.502}{0.005}$ $Z = \frac{x - μ}{σ} = \frac{x - 0.502}{0.005}$

Probability that a washer is not defective

$\begin{array}{l} = P (0.496 \leq X \leq 0.508) \\ = P (\frac{0.496 - 0.502}{0.005} \leq \frac{X - 0.502}{0.005} \leq \frac{0.508 - 0.502}{0.005}) \\ = P (- 1.2 \leq X \leq 1.2) \\ = 2 \cdot P (0 \leq X \leq 1.2) (By symmetry) \\ = 2 \cdot (0.3849) = 0.7698 \end{array}$ $\begin{array}{l} = P (0.496 \leq X \leq 0.508) \\ = P (\frac{0.496 - 0.502}{0.005} \leq \frac{X - 0.502}{0.005} \leq \frac{0.508 - 0.502}{0.005}) \\ = P (- 1.2 \leq X \leq 1.2) \\ = 2 \cdot P (0 \leq X \leq 1.2) (By symmetry) \\ = 2 \cdot (0.3849) = 0.7698 \end{array}$

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Table of Contents for 4.22 Discrete Uniform Distribution

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4.22 Discrete Uniform Distribution

4.23 The Negative Binomial and Geometric Distribution

4.24 Geometric Distribution

4.25 Continuous Probability Distributions

4.25.1 Uniform Distribution

4.25.1.1 Moments of the Uniform Distribution

4.25.1.2 Mean of Uniform Distribution

4.25.1.3 Variance of Uniform Distribution

4.25.1.4 Moment Generating Function of the Uniform Distribution

4.25.2 Exponential and Negative Exponential Distribution

4.26 Normal Distribution

4.26.1 Standard Normal Variable

4.26.2 Distribution Function φ(Z)φ(Z)<math><mi>φ</mi><mo stretchy="false">(</mo><mi>Z</mi><mo stretchy="false">)</mo></math> of Standard Normal Variate

4.26.3 Area Under Normal Curve

4.26.4 Area Under Standard Normal Curve

4.26.5 Properties of Normal Curve

4.26.6 Mean of Normal Distribution

4.26.7 Variance of Normal Distribution

4.26.8 Mode of Normal Distribution

4.26.9 Median of the Normal Distribution

4.26.10 Moment Generating Function of Normal Distribution With Respect to Origin

4.26.11 Mean Deviation of Normal Distribution

4.26.11.1 Solved Examples

Table of Contents for
4.22 Discrete Uniform Distribution

4.26.2 Distribution Function $φ (Z)$ $φ (Z)$ of Standard Normal Variate