Normal distribution

The probability density function of the random variable $X$ is given by

$$g(x) = \frac{1}{\sqrt{2\pi} \sigma} EXP\left[-\frac{(x-\mu)^2}{2 \sigma^2}\right], \ -\infty < x < \infty$$

and we say that the random variable $X$ follows the normal distribution and write $X \sim N(\mu, \sigma^2)$．

Also，the value of

$$P_{r}(0 \leq Z \leq z) = \frac{1}{\sqrt{2\pi}}\int_{0}^{z}e^{-\frac{x^{2}}{2}}dx$$

is given as standard normal table.

Normalization

Converting the mean $E(X)$ of the random variable $X$ to 0 and the variance $V(X)$ to 1 is called normalization. Normalization method Let

$$Z = \frac{X - E(X)}{\sqrt{V(X)}}$$

Then

$$E(Z) = 0, \ V(Z) = 1$$

Exercise10

1. For $X \sim N(170,10^2)$, find the following probabilities.．

(a)

$P_{r}(X \leq 160)$

(b)

$P_{r}(160 \leq X \leq 175)$

2. For $Z \sim N(0,1)$, find the $\lambda$ satisfying the followings．

(a)

$P_{r}(Z > \lambda) = 0.05$

(b)

$P_{r}(\vert Z\vert > \lambda) = 0.05$

3. The height of boys at the age of 20 nationwide shall follow the normal distribution $N(168.9,5.6^2)$.

(a)

How much should the boundary value be in order to divide the total number into 10 equal parts in order of height.

(b)

Extract 120 20-year-old boys and find the probability that the average height depends on 1.3 cm or more from 168.9 cm.．

4. The following can be said about the binomial distribution, Poisson distribution, and normal distribution.
For $X \sim B(n,p)$，

$$X \sim \left\{\begin{array}{cl} P_{o}(\mu) &, np \leq 5\\ N(\mu,\sigma^2)&, np > 5 \end{array}\right. \$$

Using this, answer the following questions．

(a)

For $X \sim B(100,0.02)$，find $P_{r}(X \geq 2)$．

(b)

For $X \sim B(100, 0.2)$，find $P_{r}(X \geq 25)$．

(c)

Throw one die 600 times and approximate the probability that number 1 rollos $S$ is 90 or more and 100 or less.

(d)

Throw 100 unbiased coins at the same time. Approximate the probability that the number of head appearances is 40 or more and 60 or less.．

Answer

1. Since $X \sim N(170,10^2)$，let $Z = \frac{X - 170}{10}$. then $Z \sim N(0,1)$.

$$P_{r}(X \leq 160) = P_{r}(\frac{X - 170}{10} \leq \frac{160 - 170}{10}) = P_{r}(Z \leq -1)$$

$$= 0.5 - P_{r}(0 \leq Z \leq 1) = 0.5 - 0.3413 = 0.1587$$

$$P_{r}(160 \leq X \leq 175) = P_{r}(\frac{160 - 170}{10} \leq \frac{X - 170}{10} \leq \frac{175 - 170}{10}) = P_{r}(-1 \leq Z \leq 0.5)$$

$$= P_{r}(-1 \leq Z \leq 0) + P_{r}(0 \leq Z \leq 0.5) = 0.3413 + 0.1915 = 0.5328$$

2.

Find $\lambda$ satisfying $P_{r}(\vert Z\vert > \lambda) = 0.05$. According to the standard normal distribution table, the value of $P_{r}(0 \leq Z \leq z)$ is given. Then we write

$$P_{r}(\vert Z\vert > \lambda) = 1 - P_{r}(\vert Z\vert \leq \lambda)$$

$$P_{r}(0 \leq Z \leq \lambda) = \frac{1}{2}(1 - P_{r}(\vert Z\vert > \lambda)) = 0.225$$

Therefore, ， $\lambda = 1.645$.

Find the $\lambda$ so that $P_{r}(\vert Z\vert > \lambda) = 0.05$. According to the standard normal distribution table, we can find the value of $P_{r}(0 \leq Z \leq z)$. Rewrite

$$P_{r}(Z > \lambda) = 0.5 - P_{r}(0 \leq Z \leq \lambda)$$

as

$$P_{r}(0 \leq Z \leq \lambda) = 0.5 - P_{r}(Z > \lambda) = 0.45$$

Then， $\lambda = 0.595$.

3.

Normalized from mean $168.9$, variance $5.6^{2}$，

$$Z = \frac{X - 168.9}{5.6}$$

Then by the standard normal distribution table, we find $z$ so that

$$P_{r}(0 \leq Z \leq z) = 0.1, \ P_{r}(0 \leq Z \leq z) = 0.2, \ldots, P_{r}(0 \leq Z \leq z) = 0.5$$

，

$$P_{r}(0 \leq Z \leq 0.255) = 0.1$$

$$P_{r}(0 \leq Z \leq 0.525) = 0.2$$

$$P_{r}(0 \leq Z \leq 0.845) = 0.3$$

$$P_{r}(0 \leq Z \leq 1.281) = 0.4$$

$$P_{r}(0 \leq Z \leq 3.4) = 0.5$$

Thus, To divide the total number into 10 equal parts, divide the value of $Z$ as follows.

$$-3.4 \sim -1.281 \sim -0.845 \sim -0.525 \sim -0.255 \sim 0 \sim 0.255 \sim 0.525 \sim 0.845 \sim 1.281 \sim 3.4$$

Let $\bar{X}$ be the average height of 120 persons. Thenby the central limit theorem,

$$\bar{X} \sim N(168.9, \frac{5.6^{2}}{120})$$

Now the probability of the average height is 1.3cm or more than 168.9cm is

$$P_{r}(\vert\bar{X} - 168.9\vert \geq 1.3) = 1 - 2P_{r}(0 \leq \bar{X} - 168.9 < 1.3)$$

Note that，

$$P_{r}(0 \leq \bar{X} - 168.9 < 1.3) = P_{r}(0 \leq \frac{\bar{X} ... ...rt{120}} \leq \frac{1.3}{5.6/\sqrt{120}}) = P_{r}(0 \leq Z \leq 2.543) = 0.4945$$

Thus

$$P_{r}(\vert\bar{X} - 168.9\vert \geq 1.3) = 1 - 2(0.4945) = 0.011$$

4.

Since $X \sim B(100,0.02)$， $E(X) = np = 2$. Then we can approximate with the Poisson distribution. Note that $X \sim P_{o}(\lambda)$， $\lambda = E(X) = 2$. Then

$$P_{r}(X \geq 2) = 1 - P_{r}(X < 2) = 1 - P_{r}(X = 0) - P_{r}(X = 1)$$

$$= 1 - P_{o}(0) - P_{o}(1) = 1 - \frac{2^{0}}{0!}e^{-2} - \frac{2^{1}}{1}e^{-2}= 0.594$$

Since $X \sim B(100, 0.2)$， $E(X) = np = 20$. Then we approximae using the normal distribution. $\mu = E(X) = np = 20, \sigma^{2} = V(X) = np(1-p) = 16$ implies $X \sim N(20,4^{2})$. Thus

$$P_{r}(X \geq 25) = 1 - P_{r}(0 \leq X < 25) = 1 - P_{r}(0 \leq \frac{X - 20}{4} < \frac{25-20}{4} )$$

$$= 1 - P_{r}(0 \leq Z < 1.25) = 1 - 0.3944 = 0.6056$$

Let $X$ be the number of 1 rolls. Then $X \sim B(600,1/6)$. Then find the probability that $S$ is 90 or more and 100 or less.

$$P_{r}(90 \leq X \leq 100) = P_{r}(X = 90) + P_{r}(X = 91) + \cdots + P_{r}(X = 100)$$

Since $\mu = E(X) = np = 100$, $\sigma^{2} = V(X) = npq = 500/6$， $X \sim N(100, 500/6)$. Using the standard distribution,

$$P_{r}(90 \leq X \leq 100) = P_{r}(\frac{90 - 100}{\sqrt{500/6}} \leq Z \leq 0)= P_{r}(0 \leq Z \leq \sqrt{1.2})$$

$$= P_{r}(0 \leq Z \leq 1.095) = 0.3621$$

If $X$ is the number of times the eye head appears, it follows the binomial distribution of $X \sim B(100,1/2)$. Therefore, if the probability that the number of times the head comes out is 40 or more and 60 or less is calculated,

$$P_{r}(40 \leq X leq 60) = P_{r}(X = 40) + P_{r}(X = 41) + \cdots + P_{r}(X = 60)$$

Since $\mu = E(X) = np = 50$, $\sigma^{2} = V(X) = npq = 25$， $X \sim N(50, 5^{2})$. Using the standard distribution,

$$P_{r}(40 \leq X \leq 60) = P_{r}(\frac{40 - 50}{5} \leq Z \leq \frac{60 - 50}{5})= 2P_{r}(0 \leq Z \leq 2)$$

$$= 2(.4772) = 0.9544$$