Chapter 6, Problem 43E

a)

To determine

To draw the model which shows 68-95-99.7

Explanation of Solution

Given:

  $\mu$ = 188

  $\sigma$ = 24s

Following is the model which shows 68-95-99.7:

  

b)

To determine

To find the percent of adult women which have cholesterol levels over 200 mg/dl

Answer to Problem 43E

The 30.85% of adult women which have cholesterol levels over 200 mg/dl.

Explanation of Solution

Given:

  $\mu$ = 188

  $\sigma$ = 24

Formula:

  $z=\frac{x-\mu }{\sigma }$

Using z-score,

  $z=\frac{200-188}{24}=0.5$

  $\begin{array}{l}P(Z>0.5)=1-P(Z<0.5)\\ P(Z>0.5)=1-0.6915=0.3085\end{array}$

Hence, 30.85% of adult women which have cholesterol levels over 200 mg/dl.

c)

To determine

To find the percent of adult women which have cholesterol levels between 150 and 170 mg/dl

Answer to Problem 43E

The 17% of adult women which have cholesterol levels between 150 and 170 mg/dl

Explanation of Solution

Given:

  $\mu$ = 188

  $\sigma$ = 24

Formula:

  $z=\frac{x-\mu }{\sigma }$

Using z-score,

  $\begin{array}{l}{z}_1=\frac{150-188}{24}=-1.5833\\ z_2=\frac{170-188}{24}=-0.75\end{array}$

  $\begin{array}{l}P(-1.5833<Z<-0.75)=P(Z<-0.75)-P(Z<-1.5833)\\ P(-1.5833<Z<-0.75)=0.2266-0.0567=0.17\end{array}$

Hence, 17% of adult women which have cholesterol levels between 150 and 170 mg/dl.

d)

To determine

To find IQR.

Answer to Problem 43E

The IQR is 32.16

Explanation of Solution

Given:

  $\mu$ = 188

  $\sigma$ = 24

Formula:

  $z=\frac{x-\mu }{\sigma }$

  $IQR=Q_3-Q_1$

First, we need to find first quartile and third quartile. Therefore, we need to find z-score corresponding to a probability of 0.25 or 0.75.

  z = $\pm 0.67$

Using formula,

  $\begin{array}{l}{Q}_1=188+(-0.67)\times 24=171.92\\ Q_3=188+0.67\times 24=204.08\end{array}$

So, $IQR=204.08-171.92=32.16$

e)

To determine

To find the highest 15% of women’s cholesterol levels.

Answer to Problem 43E

The highest 15% of women’s cholesterol levels is 212.96 mg/dL

Explanation of Solution

Given:

  $\mu$ = 188

  $\sigma$ = 24

Formula:

  $x=\mu +z\sigma$

The cutoff value of the highest 15% of the weights is same as the lowest 85% of the weights.

Therefore,

  P(Z< z) = 0.85

Z = 1.04

Using formula,

  $x=188+1.04\times 24=212.96$

Hence, the highest 15% of women’s cholesterol levels is 212.96 mg/dL.