Chapter 6, Problem 45E

a)

To determine

To find fraction of kindergarten kids should the company expect to be less than 3 feet tall.

Answer to Problem 45E

The 11.12% of kindergarten kids should the company expected to be less than 3 feet tall.

Explanation of Solution

Given:

  $\mu$ = 38.2

  $\sigma$ = 1.8

  $x$ =36

Using z-score,

  $z=\frac{36-38.2}{1.8}=-1.22$

  $P(Z<-1.22)=0.1112=11.12\%$

Hence, 11.12% of kindergarten kids should the company expected to be less than 3 feet tall.

b)

To determine

To find the percent of middle 80% of kindergarteners.

Answer to Problem 45E

The middle 80%of kindergarteners are expected to be between 35.896 and 40.504 inches.

Explanation of Solution

Given:

  $\mu$ = 38.2

  $\sigma$ = 1.8

Middle 80% values mean we need to find z-scores corresponding to a probability of 0.1 or 0.9.

  P(-z

So, z = $\pm 1.28$

Therefore,

  $\begin{array}{l}x=38.2+(-1.28)\times 1.8=35.896\\ x=38.2+1.28\times 1.8=40.504\end{array}$

Hence, the middle 80% of kindergarteners are expected to be between 35.896 and 40.504 inches.

c)

To determine

To find the biggest 10% of kindergarteners.

Answer to Problem 45E

The biggest 10% of kindergarteners are expected to be more than 40.504 inches tall.

Explanation of Solution

Given:

  $\mu$ = 38.2

  $\sigma$ = 1.8

The biggest 10% means z corresponding to probability of 90%

  P(Z< z) = 0.90

So, z = 1.28

Therefore,

  $x=38.2+1.28\times 1.8=40.504$

Hence, the biggest10% of kindergarteners are expected to be more than 40.504 inches tall.