4.26.12 Fitting a Normal Distribution

There are two methods for fitting normal distribution, namely

We explain these methods with the help of examples.

Example 4.95: Fit a normal distribution for the following data by the ordinates method:

Solution: In order to fit a normal distribution we compute the values of mean (σ) and SD (σ) of the distribution using the values of (of μ and σ) obtained, we compute

$Z=\frac{x-\mu }{\sigma }$

For the given values of x, after finding the values of Z and using the table of areas the values of φ(Z) are obtained. The ordinate Z for any value of φ(Z) is obtained by the relation

$Z=\frac{N}{\sigma }\phi (Z)$

The theoretical frequencies can be calculated by using the above relation

Class interval	f	xi(midpoint)	fixi	$f_i{x}_i^2$
150–160	9	155	1345	216,225
160–170	24	165	3960	653,400
170–180	51	175	8925	156,187
180–190	66	185	12210	2,258,850
190–200	72	195	14040	2,737,800
200–210	48	205	9840	2,017,200
210–220	21	215	4515	970,725
220–230	6	225	1350	303,750
230–240	3	235	705	165,675
	300		56940	10,885,500

$\begin{array}{ll}\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}\mathrm{is}\mu \hfill & =\frac{\sum x_i{f}_i}{\sum f_i}\hfill \\ \hfill & =\frac{56940}{300}=189.8\hfill \end{array}$

$\begin{array}{ll}\sigma =\mathrm{Standard}\mathrm{deviation}\hfill & =\sqrt{\frac{\sum f_i{x}_i^2}{\sum f_i}-{\mu }^2}\hfill \\ \hfill & =\sqrt{\frac{10885500}{300}-{189.8}^2}=16.15\hfill \end{array}$

Hence σ=16.15

Using the values of μ and σ we fit the normal distribution for the given data as follows (where C denotes the length of the class interval).

Class interval	Midpoint, x	$\left|Z\mathbf{=}\frac{x\mathbf{–}\mu }{\sigma }\right|$	Value of $\frac{N}{\sigma }\phi (Z)$	$\left[\frac{N}{\sigma }\phi (Z)\right]\mathbf{\times }\mathit{C}$	The ordinate $\phi (Z)$
150–160	155	2.15	0.0369	0.7356	7.35~7
160–170	165	0.5356	0.1238	2.299	28.99~23
170–180	175	0.9164	0.2637	4.902	49
180–190	185	0.297	0.3825	7.1113	71
190–200	195	0.322	0.379	7.05	71
200–210	205	0.942	0.2565	4.77	48
210–220	215	1.562	0.1182	2.198	22
220–230	225	2.181	0.0371	0.689	7
230–240	235	2.80	0.0079	0.15	2

Hence the expected frequencies are 7, 23, 49, 71, 71, 48, 22, 7, and 2.

Example 4.96: Fit a normal distribution to the following data and calculate the theoretical frequencies by area method:

Solution:

We have

$\sum f_i=N=100$

$\sum f_i{x}_i=6745$

$\sum f_i{x}_i^2=455,803$

$\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}\mathrm{is}\overline{x}=\frac{\sum x_i{f}_i}{\sum f_i}=\frac{6745}{100}=67.45$

$\begin{array}{ll}\sigma =\mathrm{Standard}\mathrm{deviation}\hfill & =\sqrt{\frac{\sum f_i{x}_i^2}{\sum f_i}-{\mu }^2}\hfill \\ \hfill & =\sqrt{\frac{455803}{100}-{(67.45)}^2}=2.92\hfill \end{array}$

Hence σ=2.92

The expected frequencies can be computed as shown in the table given below. Hence the theoretical frequencies are 4, 21, 39, 28, and 7.

Lower limits	$Z=\frac{x-\mu }{\sigma }$	Area under the normal curve $\phi (Z)$	Area included in each class $\mathit{\Delta }\mathit{\phi }\mathit{(}Z\mathit{)}$	Expected frequency $\mathit{N}\cdot \mathit{\Delta }\mathit{\phi }\mathit{(}Z\mathit{)}$
59.2	−2.72	0.4967	0.0413	4.13~4
62.5	−1.70	0.4554	0.2068	2068~21
65.5	−0.67	0.2486	0.3892	38.92~39
68.5	0.36	0.1406	0.2771	27.71~28
71.5	1.39	0.4177	0.0743	7.43~7
74.5	2.4	0.4920

Example 4.97: Find the points of inflection of a normal curve. Find the asymptote of the cure?

Solution: The equation of the normal curve is

$y=f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}$ (4.24)

(4.24)

Differentiating Eq. (4.24) with respect to x, we get

$\begin{array}{ll}f\prime (x)\hfill & =\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}\left[-\frac{\left(x-\mu \right)}{{\sigma }^2}\right]\hfill \\ \hfill & =-\frac{1}{\sigma \sqrt{2\pi }{\sigma }^2}(x-\mu ).e^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}\hfill \end{array}$

Again differentiating we get

$f″(x)=\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{1}{\sigma \sqrt{2\pi }}\left[e^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}{\left[-\frac{(x-\mu )}{{\sigma }^2}\right]}^2-\frac{1}{{\sigma }^2}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\right]=0$

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=0\mathrm{when}{\left(\frac{x-\mu }{\sigma }\right)}^2-1=0$

i.e.,

${(x-\mu )}^2={\sigma }^2$

or

$(x-\mu )=\pm \sigma$

or

$x=\mu \pm \sigma$

Differentiating $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ with respect to x, we get

$\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}=\frac{1}{\sigma \sqrt{2\pi }}\left[e^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}{\left[-\frac{(x-\mu )}{{\sigma }^2}\right]}^3+\frac{3(x-\mu )}{{\sigma }^2}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}\right]$

Hence $\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}=\pm \frac{2}{{\sigma }^4\sqrt{2\pi }}{e}^{-\frac{1}{2}}$ when $\mu =x\pm \sigma$

i.e., $\frac{{\mathrm{d}}^{3}y}{\mathrm{d}{x}^3}\ne 0$ when $\mu =x\pm \sigma$

The values of y=f(x) at $x=\mu \pm \sigma$ is $\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}}$

Therefore the points of inflection are $\left(\mu \pm \sigma ,\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}}\right)$

The line y=0, meets the curve at x=±∞.

Hence the curve has x-axis for its asymptote.

Exercise 4.10

4.26.13 Linear Combination of Independent Normal Variables

Proof

Since

$M_x(t)=e^{t\mu +\frac{1}{2}{t}^2{\sigma }^2}$

We have

$\begin{array}{ll}M{(t)}_{a_i{x}_i}\hfill & =M_X(a_{i}t)=e^{a_i{\mu }_i+\frac{1}{2}{a_i}^2{t}^2{{\sigma }_i}^2}\hfill \\ \hfill & =e^{t(a_i{\mu }_i)+\frac{1}{2}{t}^2{(a_i{\sigma }_i)}^2}\hfill \end{array}$

$\begin{array}{ll}M{(t)}_{\sum a_i{x}_i}\hfill & =M{(t)}_{a_1{x}_1+a_2{x}_2+\cdots +a_n{x}_n}\hfill \\ \hfill & =M_{a_1{x}_1}(t)\cdot M_{a_2{x}_2}(t)\dots M_{a_n{x}_n}(t)\hfill \\ \hfill & =e^{t(a_1{\mu }_1)+\frac{1}{2}{t}^2{(a_1{\sigma }_1)}^2}\cdot e^{t(a_2{\mu }_2)+\frac{1}{2}{t}^2{(a_2{\sigma }_2)}^2}\dots e^{t(a_n{\mu }_n)+\frac{1}{2}{t}^2{(a_n{\sigma }_n)}^2}\hfill \\ \hfill & =e^{t\sum _1^n(a_i{\mu }_i)+\frac{1}{2}{t}^2\sum _1^n{(a_i{\sigma }_i)}^2}\hfill \end{array}$

is the mgf of a normal variate with mean $\sum a_i{\mu }_i$ and variance, $\sum a_i^2{\sigma }_i^2$.

Hence the theorem.

4.26.14 Fitting a Normal Distribution

If N denotes the sum of all frequencies of a normal distribution, the normal curve is given by $y=\frac{N}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}$.

The area under the normal curve is equal to N.

To fit a normal curve to a distribution, there are two methods namely

These methods are explained with the help of few examples.

Example 4.98: Fit a normal curve to the distribution

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

We have,

$N=\sum f_i=1+2+3+2+1=9$

$\begin{array}{ll}\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}\mathrm{is}\overline{x}\hfill & =\frac{\sum x_i{f}_i}{\sum f_i}=\frac{1.1+3.2+5.3+7.2+9.1}{9}\hfill \\ \hfill & =\frac{45}{9}=5\hfill \end{array}$

$\begin{array}{ll}{\sigma }^2=\mathrm{variance}\hfill & =\frac{\sum {x_i}^2{f}_i}{\sum f_i}-{\mu }^2\hfill \\ \hfill & =\frac{1^2\cdot 1+3^2\cdot 2+5^2\cdot 3+7^2\cdot 2+9^2\cdot 1}{9}-5^2\hfill \\ \hfill & =\frac{1+18+75+98+81}{9}-25\hfill \\ \hfill & =\frac{273}{9}-25\hfill \\ \hfill & =\frac{273-225}{9}=\frac{48}{9}\hfill \\ \therefore \sigma \hfill & =\sqrt{\frac{48}{9}}=\frac{4\sqrt{3}}{3}\hfill \end{array}$

The equation of best fitting curve is

$\begin{array}{ll}y\hfill & =\frac{N}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}\hfill \\ \hfill & =\frac{9}{\frac{4\sqrt{3}}{3}\sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-5}{\frac{48}{9}}\right]}^2}\hfill \\ \hfill & =\frac{9\sqrt{3}}{4\sqrt{2\pi }}{e}^{-\frac{9}{96}{[x-5]}^2}\hfill \\ \hfill & =\frac{9\sqrt{3}}{4\sqrt{2\pi }}{e}^{-\frac{3}{32}{[x-5]}^2}\hfill \end{array}$

Example 4.99: Fit a normal distribution to the following data by the method of ordinates:

Solution:

We have

$\sum f_i=N=100$

$\sum f_i{x}_i=6745$

$\sum f_i{x}_i^2=455,803$

$\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}\mathrm{is}\overline{x}=\frac{\sum x_i{f}_i}{\sum f_i}=\frac{6745}{100}=67.45$

$\begin{array}{ll}\sigma =\mathrm{Standard}\mathrm{deviation}\hfill & =\sqrt{\frac{\sum f_i{x_i}^2}{\sum f_i}-{\mu }^2}\hfill \\ \hfill & =\sqrt{\frac{455803}{100}-{(67.45)}^2}=2.92\hfill \end{array}$

Hence σ=2.92

The ordinates are obtained at various distances from the mean, we use the formula

$f(x)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}$

where

$Z=\frac{x-\mu }{\sigma }$

We have the following table:

Class interval	x(Midpoint)	$Z\mathit{=}\frac{\mathit{x}-\mathit{\mu }}{\mathit{\sigma }}$	$f(x)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}$	$y=\frac{N}{\sigma }f(Z)$	$\frac{N}{\sigma }f(Z)\times c$
59.5–62.5	61	−2.21	0.0347	1.188	3.565
62.5–65.5	64	−1.18	0.1984	6.794	20.382
65.5–68.5	67	−0.15	0.3943	13.50	40.50
68.5–71.5	70	0.87	0.2725	9.333	27.999
71.5–74.5	73	1.90	0.0656	2.247	6.741

The expected (i.e., theoretical) frequencies are 3.565, 20.382, 40.50, 27.999, 6.741 i.e., 4, 20, 41, 28, and 7.

Example 4.100: Fit a normal distribution to the following data by the area method:

Solution: We find the values of mean and SD as follows:

We have

$\sum f_i=N=1000$

$\sum f_i{x}_i=79945$

$\sum f_i{x}_i^2=64200850.0$

$\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{distribution}\mathrm{is}\mu =\frac{\sum x_i{f}_i}{\sum f_i}=\frac{79945}{1000}=79.945$

$\begin{array}{ll}{\sigma }^2=\mathrm{variance}\hfill & =\frac{\sum x_i^2{f}_i}{\sum f_i}-{\mu }^2\hfill \\ \hfill & =\frac{64200850}{1000}-{(79.945)}^2\hfill \\ \hfill & =6420.85–639120\hfill \\ \hfill & =29.65\hfill \\ \therefore \sigma \hfill & =\sqrt{29.65}=5.4451\hfill \end{array}$

Since the normal distribution takes values from, $-\infty$ to $\infty$ the given data can be modified as follows:

Class interval	f	Lower limit	$Z=\frac{x–\mu }{\sigma }$	Area $\phi (Z)$	$\mathrm{\Delta }\phi (X)=\phi (Z+1)–\phi (Z)$	$N\cdot \mathrm{\Delta }\phi (Z)$
$–\infty$ to 60	0	$–\infty$	$–\infty$	0	0.0001	0.1
60–65	3	60	−3.6629	0.0001	0.003	3
65–70	21	62	−2.7447	0.0031	0.0313	31.3
70–75	150	70	−1.8264	0.0344	0.1497	149.7
75–80	335	75	−0.9081	0.1841	0.3199	319.9
80–85	326	80	0.0101	0.5040	0.3172	317.2
85–90	135	85	0.9283	0.8212	0.1459	145.9
90–95	26	90	1.8466	0.9671	0.03	30
95–100	4	95	2.7649	0.9671	0.0028	2.8
100	$–\infty$	$\infty$	100	3.6831	0.9999

The expected frequencies are 0.1, 3, 31.3, 149.7, 319.9, 317.2, 145.9, 30, 2.8, i.e., 0, 3, 31, 150, 320, 317, 146, 30, 3.

4.26.15 Normal Approximation to Binomial Distribution

Normal can be used to approximate the binomial distribution. Consider a random variable X. If X follows binomial distribution then we have

$P(X=x)={}^{n}C_x{p}^x{q}^{n-x}$

which is very large, then

$P(X_1<x<x_2)=\sum _{x=x_1}^{n=x_2}{}^{n}C_x{p}^x{q}^{n-x}$

Example 4.101: Find the probability of getting 1 or 3 or 4 or 5 in throwing a die 5–7 times among 9 trials?

Solution: We have n=9

$p=4\cdot \frac{1}{6}=\frac{2}{3}$

$q=1–p=1–\frac{2}{3}=\frac{1}{3}$

Mean $\sigma =np=9$·$\frac{2}{3}=6$

${\sigma }^2=\mathrm{variance}=npq=9·\frac{2}{3}·\frac{1}{3}=2$

$\therefore \sigma =\sqrt{2}=1.414$

$x_1=5,x_2=7$

$\begin{array}{ll}{Z}_1\hfill & =\frac{x_1-\frac{1}{2}-\mu }{\sigma }=\frac{5-\frac{1}{2}-6}{1.414}=\frac{4.5-6}{1.414}\hfill \\ \hfill & =\frac{-1.5}{1.414}=-1.0608\hfill \end{array}$

$\begin{array}{ll}{Z}_2\hfill & =\frac{x_2-\frac{1}{2}-\mu }{\sigma }=\frac{7-\frac{1}{2}-6}{1.414}=\frac{7.5-6}{1.414}\hfill \\ \hfill & =\frac{1.5}{1.414}=1.0608\hfill \end{array}$

$P(X_1<X<X_2)=P\left(X_1-\frac{1}{2}<X<X_2+\frac{1}{2}\right)$

$\begin{array}{ll}\therefore P(4.5<X<7.5)\hfill & =P(-1.0608<Z<1.0608)\hfill \\ \hfill & =P(-1.0608<Z<0)+P(0<Z<1.0608)\hfill \\ \hfill & =2[P(0<Z<1.0608)]=2(0.3554)\hfill \\ \hfill & =0.7108\hfill \end{array}$

Example 4.102: Eight coins are tossed together, find the probability of getting 1–5 heads in a single toss?

Solution: We have n=8

$p=\frac{1}{2}\mathrm{and}q=\frac{1}{2}$

Mean $\mu =np=8$·$\frac{1}{2}=4$

${\sigma }^2=\mathrm{variance}=npq=8.\frac{1}{2}\cdot \frac{1}{2}=2$

$\therefore \sigma =\sqrt{2}=1.414$

$x_1=1,x_2=5$

$\begin{array}{ll}Z\hfill & =\frac{x_1-\frac{1}{2}-\mu }{\sigma }=\frac{1-\frac{1}{2}-4}{1.414}\hfill \\ \hfill & =\frac{-3.5}{1.414}=-2.4752\hfill \end{array}$

$\begin{array}{ll}{Z}_2\hfill & =\frac{x_2-\frac{1}{2}-\mu }{\sigma }=\frac{5-\frac{1}{2}-4}{1.414}\hfill \\ \hfill & =\frac{3.5}{1.414}=2.4752\hfill \end{array}$

$\begin{array}{ll}\therefore P(1<X<5)\hfill & =P(-2.475<Z<2.475)\hfill \\ \hfill & =2[P(0<Z<2.475)]=2(0.4932)\hfill \\ \hfill & =0.9864\hfill \end{array}$

Example 4.103: Random variable X is normally distributed with m=12 and SD 2. Find P(9.6<x<13.8)?

Given that for $\frac{x}{\sigma }=0.9$, A=0.3159 and

for $\frac{x}{\sigma }=1.2$ A=0.3849

Solution: We have

$P(9.6<x<13.8)={\int }_{9.6}^{13.8}\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-\mu }{\sigma }\right]}^2}\mathrm{d}x$ (4.25)

(4.25)

$\mu =12,\sigma =2,Z=\frac{x-\mu }{\sigma }=\frac{x-12}{2}$

When

$X=9.6,Z=Z_1=\frac{9.6-12}{2}=-1.2$

and

$X=13.8,Z=Z_2=\frac{13.8-12}{2}=0.9$

Substituting in Eq. (4.25) we get

$\begin{array}{ll}P(9.6<x<13.8)\hfill & ={\int }_{9.6}^{13.8}\frac{1}{2\sqrt{2\pi }}{e}^{-\frac{1}{2}{\left[\frac{x-12}{2.4}\right]}^2}\mathrm{d}x\hfill \\ \hfill & =\frac{1}{\sqrt{2\pi }}{\int }_{-1.2}^{0.9}{e}^{-\frac{1}{2}{z}^2}\mathrm{d}x\hfill \\ \hfill & =\frac{1}{\sqrt{2\pi }}{\int }_{-1.2}^0{e}^{-\frac{1}{2}{z}^2}\mathrm{d}x+\frac{1}{\sqrt{2\pi }}{\int }_0^{0.9}{e}^{-\frac{1}{2}{z}^2}\mathrm{d}x\hfill \\ \hfill & =0.3849+0.3159=0.7008\hfill \end{array}$

4.28.1 Mean and Variance of Gamma Distribution

Since

$M_X(t)={(1-t)}^{-\lambda }$

We get

$\begin{array}{ll}{M\prime }_x(t)\hfill & =\lambda {(1-t)}^{-\lambda -1}(-1)\hfill \\ \hfill & =\lambda {(1-t)}^{-\lambda -1}\hfill \end{array}$

Hence

${\mu \prime }_1={M\prime }_x(0)=\lambda$

Since

${M\prime }_x(t)=\lambda {(1-t)}^{-\lambda -1},\mathrm{we}\mathrm{get}$

$\begin{array}{ll}{M″}_X(t)\hfill & =(-\lambda -1)\lambda {(1-t)}^{-\lambda -2}(-1)\hfill \\ \hfill & =(\lambda +1)\lambda {(1-t)}^{-\lambda -2}\hfill \\ \therefore {\mu \prime }_2\hfill & ={M″}_X(0)=\lambda (\lambda +1)\hfill \end{array}$

$\begin{array}{ll}\mathrm{Variance}\hfill & ={\mu \prime }_2-{\mu \prime }_2\hfill \\ \hfill & =\lambda (\lambda +1)-{\lambda }^2\hfill \\ \hfill & ={\lambda }^2+\lambda -{\lambda }^2=\lambda \hfill \end{array}$

Hence mean of Gamma distribution=variance of gamma distribution=λ

The first four central moments of Gamma distribution of first kind are

${\mu }_1=0,{\mu }_2=\lambda ,{\mu }_3=2\lambda \mathrm{and}{\mu }_4=3\lambda (\lambda +2)$

4.28.2 Gamma Distribution of Second Kind

A random variable x is said to follow Gamma distribution of second kind if it has the following density function:

$\begin{array}{lll}f(x)=P(X=x)\hfill & =\frac{a^{\lambda }}{\mathrm{\Gamma }\lambda }{e}^{-ax}{x}^{\lambda -1},\hfill & a>0,\lambda >0,0<x<\infty \hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

4.29.1 Beta Distribution of Second Kind

A continuous random variable x is said to follow beta distribution of second kind if its pdf is given by

$\begin{array}{lll}f(x)\hfill & =\frac{1}{\beta (m,n)}\cdot \frac{x^{m-1}}{{(1+x)}^{m+n}},\hfill & m,n>0,0<x<\infty \hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$