91Ó°ÊÓ

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,

$p(65<X<69)=p\left(Z≤\frac{X_2−\mathrm{μ}}{\mathrm{σ}}\right)−p\left(Z≤\frac{X_1−\mathrm{μ}}{\mathrm{σ}}\right)$

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

$=p(Z≤1.2)−p(Z≤−0.4)$

The calculation of $p(Z≤1.2)$and $p(Z≤−0.4)$is done from the Z value table.

$p(z<1â‹\dots 2)−p(z≤−0.4)$

$P(z≤1â‹\dots 2=P(−\mathrm{∞}≤z≤0)+P(z≤1â‹\dots 2)$

$=0.5+3849$

$=0.8849$.

Calculation of $P(z≤−0â‹\dots 4)$

$=P(−\mathrm{∞}≤z≤0)−P(0≤2≤0.4)$

$=0.5−0.1554$

$=0.3446$

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

$p(65<X<69)=p(Z≤1.2)−p(Z≤−0.4)$

$=0.8849−0.3446$

$=0.5403$

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,

$p(X>72)=1−\left(Z≤\frac{X−\mathrm{μ}}{\mathrm{σ}}\right)$

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

$=1−p(Z≤2.4)$

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

$P(Z≤2.4)$

$=0.500+.4918$

$=0â‹\dots 9918$

Now we need to put the value in the above,

$p(X>72)=1−p(Z≤2.4)$

$=1−0.9918$

$=0.0082$

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.