According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of $66$inches and a standard deviation of $2.5$inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual.

a.$X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

b. Find the probability that the person is between $65$and $69$inches. Include a sketch of the graph, and write a probability statement.

c. Would you expect to meet many Asian adult males over 72 inches? Explain why or why not, and justify your answer numerically.

Given in the question that

Average or mean $=66$

Standard deviation$=2.5$

$X=$height of the individual

Here we need to fond the $X~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

For that

mean is $66$

Standard deviation is $2.5$

so the values is$X~N(66,2.5)$

The normal distribution that follows the random variable $X$as$X~N(66,2.5)$.

Given in the question that

Mean $=66$

Standard deviation $=2.5$

$X=$height of the individual

The probability for the person between $65$and $69$inches is calculated below,

$p(65<X<69)=p\left(Z\le \frac{X_2-\mu }{\sigma }\right)-p\left(Z\le \frac{X_1-\mu }{\sigma }\right)$

$=p\left(Z\le \frac{69-66}{2.5}\right)-p\left(Z\le \frac{65-66}{2.5}\right)$

$=p(Z\le 1.2)-p(Z\le -0.4)$

The calculation of $p(Z\le 1.2)$and $p(Z\le -0.4)$is done from the Z value table.

$p(z<1\cdot 2)-p(z\le -0.4)$

$P(z\le 1\cdot 2=P(-\mathrm{\infty }\le z\le 0)+P(z\le 1\cdot 2)$

$=0.5+3849$

$=0.8849$.

Calculation of $P(z\le -0\cdot 4)$

$=P(-\mathrm{\infty }\le z\le 0)-P(0\le 2\le 0.4)$

$=0.5-0.1554$

$=0.3446$

Now we need to put the value in the above formula,

$p(65<X<69)=p(Z\le 1.2)-p(Z\le -0.4)$

$=0.8849-0.3446$

$=0.5403$

The graphical representation of the above calculation is shown below,

The probability of the person between $65$and $69$inches is $0.5403$and the graph is

Given in the question that

Mean$=66$

Standard deviation$=2.5$

$X=$height of the individual

The calculation for the probability of Asian adult males over 72 inches is shown below,

$p(X>72)=1-\left(Z\le \frac{X-\mu }{\sigma }\right)$

$=1-p\left(Z\le \frac{72-66}{2.5}\right)$

$=1-p(Z\le 2.4)$

The calculation $p(Z\le 2.4)$is done from the Z value table

$P(Z\le 2.4)$

$=0.500+.4918$

$=0\cdot 9918$

Now we need to put the value in the above,

$p(X>72)=1-p(Z\le 2.4)$

$=1-0.9918$

$=0.0082$

Therefore, the probability that an Asian male is over $72$inches tall is $0.0082$. Since the probability that an Asian male is over $72$inches tall is $0.0082$, it is not expected that an Asian male is over $72$ inches tall.