Normal Distribution: A random variable X whose probability density function is given by
$$\phi (x)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma}}, -\infty <x< \infty ..... (1)$$
is called a normal variate. We also say
that X is normally distributed with parameters $\mu$ and $\sigma^2$, and is denoted as $X\sim N(\mu, \sigma^2)$. The continuous distribution given by eq. (1) is called the
normal distribution.
Remark 1. The cumulative distribution of $X\sim N(\mu, \sigma^2)$ is denoted as $\phi(x)$. Thus
$$ \phi(x) = P[X\leq x]=\int_{-\infty}^{x}\phi (u) du$$Remark 2. It can be verified that $\int_{-\infty}^{\infty} \phi(x)dx= 1$
Remark 3. The curve $y=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma}}$ called the normal curve which possesses the following properties:
ii. The maximum ordinate is $\frac{1}{\sigma \sqrt{2\pi}}$ , which occur at $x=\mu$
iii. Mean, median and mode of the normal distribution coincide i.e. at $x=\mu$
Standard Normal Variate: If $X\sim N(\mu,\sigma^2)$, then $Z=\frac{x-\pi}{\sigma}$ is called the standard normal variate with $ E[Z]=0 $ and $ var[z]=1$, and is written as$ Z \sim N(0,1)$.
The p.d.f. of standard normal variate Z is given by $\phi(z)=\frac{1}{\sqrt2\pi}$
and the corresponding distribution function is given by $ \phi(z)=P[Z \leq z] = \int_{-\infty}^{z} \phi (z)dz$
Remark: Since $\int_{-\infty}^{\infty} \varphi(z) dz=1$ therefore, $\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}z^2}$
Hence, $\int_{-\infty}^{\infty} e^{-\frac{1}{2} z^{2}} d z=\sqrt{2 \pi}$
Theorem: If $X \sim N\left(\mu, \sigma^{2}\right)$ , then $E[X]=\mu, var[x]=\sigma^2$ and $M_{X}(t)=e^{\mu t+\frac{1}{2} \sigma^{2} t^{2}}$
Proof: m.g.f. of X is given by $M_{X}(t)=E\left[e^{tX}\right]=\int_{-\infty}^{\infty}e^{tX}\cdot \varphi(x) dx$
$M_{X}(t)=\frac{1}{\sigma \sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{tX}\cdot e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}dx= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{t\{\mu+\sigma z\}}\cdot e^{-\frac{1}{2} z^2}dz$, where $z=\left(\frac{z-\mu}{\sigma}\right)$
$M_{X}(t)=\frac{e^{t \mu}}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}\cdot e^{-\frac{1}{2}\left(z^2-2t\sigma z \right)}dz=\frac{e^{t\mu}}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\cdot e^{-\frac{1}{2}\left((z-\sigma t)^{2}-\sigma^{2} t^{2}\right\}} d z$
$M_{X}(t)=\frac{e^{t \mu+\frac{1}{2} \sigma^2 t^2}}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \cdot e^{-frac{1}{2}(z-\sigma t)^{2}} d z . \quad(\text{ Put } u=z-\sigma t)$
$M_{X}(t)=\frac{e^{t \mu+\frac{1}{2} \sigma^{2} t^{2}}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \cdot e^{-\frac{1}{2} u^{2}} d u=e^{t \mu+\frac{1}{2} \sigma^{2} t^{2}} \times 1=e^{t \mu+\frac{1}{2} \sigma^{2} t^{2}} (1)$
Hence, $M_{X}(t)=e^{t \mu+\frac{1}{2} \sigma^{2} t^{2}}$ is the required m.g.f. of the normal distribution.
Differentiating (1) with respect to $t$
$$M_{X}^{\prime}(t)=e^{t\mu+\frac{1}{2}\sigma^2}(\mu + t \sigma^2) \Rightarrow E[X] =\left. M_{X}^{\prime}(t)\right|_{t=0}=\mu....... (2)$$
Differentiating (2) with respect to $t$
$$M_{X}^{\prime \prime} (t)=e^{t\mu +\frac{1}{2}\sigma^2 t^2} (\mu+t\sigma^2)^2 + e^{t\mu +\frac{1}{2}\sigma^2 t^2}\sigma^2$$
$$ E[X^2]=\left. M_{X}^{\prime \prime}\right|_{t=0}=\mu^2+\sigma^2............. (3)$$
Hence, $var[X]=E[X^2]-(E[X])^2=\mu^2 +\sigma^2-\mu^2=\sigma^2$
Theorem: If $X\sim N(\mu, \sigma^2),$ then $P[a<X<b]=\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)$
Theorem: If $X\sim (0, 1)$, then $P[a <X <b]=\Phi(b)-\Phi(a)=\int_a^b \Phi(z)dz$
Area under a normal curve: The definite integral xai(z) $=\int_0^z \varphi(z)dz $ is called the normal
probability integral, which gives the area under the standard normal curve
between the ordinates $Z=0$ to $Z=z$. Also due to symmetry, it gives the same area under the
ordinates $Z=-z$ to $Z=0$.
The value of $\int_0^z \varphi(z)dz=\frac{1}{\sqrt{2\pi}}\int_0^z e^-\frac{1}{2}z^2dz$ for different values of $z$
Important Results:
Example 1. If X is
a normally distributed with mean 2 and variance 1, find $P[|x-2|<1]$.
Solution: We have $\mu=2$ and $\sigma=1$. Let $Z=\frac{X-\mu}{\sigma}=X-2$
Now, $P[|X-2|<1]=P[2-1<X<2+1]=P[1<x<3]=P[1-2<X-2<3-2]$
$\Rightarrow P[|X-2|<1]=P[1-2<X-2<3-1]=P[-1<Z<1]$
$\Rightarrow P[|X-2|<1]=2P[0<Z<1]$
$\Rightarrow P[|X-2|<1]=2 \times 0.3413 = 0.6826$ from table
Example 2: If $X$ is a random variable with mean $50$ and variance $100$, find $P[Y \leq 3137]$,
Where $Y=X^2+1$ and $\xai(0.6)=0.2258$.
Solution: Given that $\mu =50$ and $\sigma=10$.
Therefore, $P[Y\leq 3137]=P[X^2+1 \leq 3137]=P[X^2+1 \leq 3137 +1]$
$$ P[Y\leq 3137]=P[X^2 \leq 3137] = P[X^2 \leq 56^2] = P[-56 \leq X \leq 56]$$
$$ P[Y \leq 3137]= P\left[\frac{-56-50}{10}\leq \frac{X-50}{10}\right]=P[-10.6 \leq Z \leq 0.6]$$
$$P[Y\leq 3137] = P=[-10.6\leq Z \leq 0] + P[0\leq Z \leq 0.6], ~by~putting~ Z=\frac{X-\mu}{\sigma}$$
$$ P[Y\leq 3137]=0.5 +\xai(0.6) = 0.5 +0.2258$$
$$ P[Y\leq 3137] = 0.7258$$
Example 3: If $log_e x$ is normally distributed with mean as unity and variance 4, find $P[\frac{1}{2} <X <2]$. Given that $log_e 2=0.693$.
Solution: $P=[\frac{1}{2} <x <2]=P[log_e \frac{1}{2} < log_e x <log_e 2]=P[-0.693<log_e x<0.693]$
$$ P[\frac{1}{2}<x<2] = P[log_e\frac{1}{2} < log_e x<log_e2]=P[-0.693<log_e x<0.693]$$
$$ P[\frac{1}{2}<x<2] =P[-0.693 < y < 0.693], ~where~ y=log_e x \sim N(1, 4)$$
$$P[\frac{1}{2}<x<2]=P[\frac{-0.693-1}{2}<\frac{y-1}{2}<\frac{0.693-1}{2}, take~ Z=\frac{y-\mu}{\sigma}=\frac{y-1}{2} \sim N(0, 1)$$
$$P\left[\frac{1}{2} <x<2 \right] = P[-0.85 < Z <-0.15] = P[0.15<Z<0.85], by ~ symmetry$$
$$P\left[\frac{1}{2} <x<2 \right] =\xai(0.85) - \xai(0.15)=0.3023- 0.0596 = 0.2427$$
Example 4. The local authorities of Chapra installed
2000 electric lamps in the street of city. If the lamps have an average life of
1000 burning hours with a S.D. of 200 hours. Assumes that the lives of the
lamps are normally distributed. (i) What
number of the lamps might be expected to fail in the first 700 burning hours, (ii) after what period of burning hours
would we expect that (a) 10% of the
lamps would have failed (b) 10% of
the lamps would be still burning?
Solution: (i) Let X denote the life of lamps, then $X\sim N(1000, 200^2)$
$$P[X < 700]=P\left[ \frac{X-1000}{200}<\frac{700-1000}{200}\right] =P[Z<-1.5],~ where ~ Z=\frac{X-\mu}{\sigma}=\frac{X-1000}{200}$$
$$ P[X<700] = P[Z>1.5]=0.5-P[0\leq Z \leq 1.5] = 0.5 - 0.433, ~ from~table$$
$$P[X<700] = 0.067$$.
Thus the number of lamps expected in first 700 hours of burning $=2000 \times 0.067=134$
ii) (a) Let $t$ denote the period in hours when 10% of the lamps would still be burning. Then
$P[X< t] = 0.1 \Rightarrow P[X \leq t] =1.0 -0.1 =0.9$
$\Rightarrow P\left[ \frac{X-1000}{200}\leq \frac{t-1000}{200} \right] = 0.9 \Rightarrow P\left[Z \leq \frac{t-1000}{200}\right]=0.9$
$\Rightarrow P \left[0 \leq Z \leq \frac{t-1000}{200}\right]=0.9-0.5=0.4=P[0\leq Z \leq 1.28]$
Hence
10% of the lamps would still be burning after 1256 burning hours.
(b) If we draw a figure, it follows from the figure that 10% of the lamps would have failed after 1000-256 = 744 burning hours
Example 5. Of a large group of men, 5% are under 60
inches in height and 40% are between 60 and 65 inches. Assuming a normal
distribution, find the mean height and standard deviation.
Solution: If $X \sim N(\mu, \sigma^2)$, then we are given $P[X<60]=0.05$ and $P[60<X<65]=0.40$
$\Rightarrow P[X < 60 ] = 0.05$ and $P[X<65]=0.05+0.40=0.45$
$\Rightarrow P\left[\frac{X-\mu}{\sigma}<\frac{60-\mu}{\sigma}\right]=0.05$ and $\Rightarrow P\left[\frac{X-\mu}{\sigma}<\frac{65-\mu}{\sigma}\right]=0.45$ (Put $Z=\frac{X-\mu}{\sigma}$)
$\Rightarrow P\left[Z<\frac{60-\mu}{\sigma}\right]=0.05$ and $\Rightarrow P\left[Z< \frac{65-\mu}{\sigma}\right]=0.45$
$\Rightarrow P\left[\frac{60-\mu}{\sigma}<Z<0 \right] = 0.5-0.05$ and $P\left[\frac{65-\mu}{\sigma}<Z<0\right]=0.5-0.45$
$\Rightarrow P\left[\frac{60-\mu}{\sigma}<Z<0 \right] = 0.45$ and $P\left[\frac{65-\mu}{\sigma}<Z<0\right]=0.05$ (due to symmetry)
$\Rightarrow P\left[0<Z<\frac{\mu-60}{\sigma}\right] = 0.45$ and $P\left[0<Z<\frac{\mu-65}{\sigma}<Z<0\right]=0.05$ (By table)
$\Rightarrow P\left[0<Z<\frac{\mu-60}{\sigma}\right]=P[0<Z<1.645]$ and $P[0<Z<\frac{\mu-65}{\sigma}=P[0<Z<0.13]$
$\Rightarrow \frac{\mu-60}{\sigma}=1.645$ and $\frac{\mu-65}{\sigma}=0.13$
$\Rightarrow \mu= 65.42$ and $\sigma = 3.29.$
Exercises:
1. X is a normal variate with mean 30 and S.D. 5. Find the probability that:
(i) $26\leq X \leq 40$, (ii) $X\geq 15$ and (iii) $|X-30|>5$
2. If $log_{10}X\sim N(4, 2^2)$ and $log_{10}1202=3.08, log_{10}8318=3.92$ Then find $P[1.202<X<83180000].$
3. Let $X$ be the life in hour of a radio tube. Assume that $X$ is normally distributed with mean 20 and variance $\sigma^2$. If a purchaser of such radio tubes requires that at least 90% of the tubes have lives exceeding 150 hours. What is the largest value of $\sigma$ can be and shall have the purchaser satisfied?
4. Suppose that the diameter of the shafts manufactured by a certain machine are normal random variable with mean 10 cms and S.D. 0.1 cm. If for a given application the shaft must meet the requirement that its diameter fall between 9.9 and 10.2 cms, what proportion of the shafts made by this machine will meet the requirement?
5. If the skulls are classified as A, B and C according as the length-breadth index is under 75, between 75 and 80, and over 80. Find approximately the mean and S. D. of a series in which A are 58%, B are 38% and C are 4%. (Assume that the distribution is normal and use table).
6. Show that for a normal distribution $\beta_1=0$ and $\beta_2=3$
7. Fit the normal curve to the data (means find $\mu, \sigma$ and normal p.d.f):
$x$ (mid-point) 2 4 6 8 10
$f$ (frequency) 1 4 6 4 1
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