Chapter 7.4, Problem 22E

Blood pressure: High blood pressure has been identified as a risk factor for heart attacks and strokes. The National Health and Nutrition Examination Survey reported that the proportion of U.S. adults with high blood pressure is 0.3. A sample of 38 U.S. adults is chosen.

1. Is it appropriate to use the normal approximation to find the probability that more than 40% of the people in the sample have high blood pressure? If so, find the probability. If not; explain why not.
2. A new sample of 80 adults is drawn. Find the probability that more than 40% of the people in this sample have high blood pressure.
3. Find the probability that the proportion of individuals in the sample of 80 who have high blood pressure is between 0.20 and 0.35.
4. Find the probability that less than 25% of the people in the sample of 80 have high blood.
5. Would it be unusual if more than 45% of the individuals in the sample of 80 had high blood pressure?

To determine

Whether it is appropriate to use the normal approximation to find the probability that more than $40\%$ of the people in the sample have high blood pressure under given scenario.

Answer to Problem 22E

The probability that more than $40\%$ of the people in the sample have high blood pressure.

is $0.0764$ .

Explanation of Solution

Given Information:

The National Health and Nutrition Examination Survey reported that the proportion of U.S. adults with high blood pressure is $\text{0}\text{.3}\text{.}$ A sample of $38$ U.S. adults is chosen.

Formula used:

The standard deviation is

  ${\sigma }_P=\sqrt{\frac{p\left(1-p\right)}{n}}$

where $n$ is sample size and $p$ is population proportion.

Calculation:

Given that the sample size is $n=38$ and population proportion is $p=0.3$ .

Since, $np=\left(38\right)\left(0.3\right)=11.4\ge 10$

And $n\left(1-p\right)=\left(38\right)\left(1-0.3\right)=26.6\ge 10$ .

This means normal approximation can be used to find the probability that more than $40\%$ of the people in the sample have high blood pressure.

The mean is ${\mu }_p=p=0.3$

The standard deviation is,

  $\begin{array}{c}{\sigma }_p=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.3\left(1-0.3\right)}{38}}\\ =0.07\end{array}$

Calculating the $z$ score for $0.40$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ =\frac{0.40-0.3}{0.07}\\ =1.43\end{array}$

The required probability is

  $\begin{array}{c}P\left(\widehat{p}>0.40\right)\\ =1-P\left(\widehat{p}<0.40\right)\\ =1-P\left(z<1.43\right)\\ =1-0.9236\\ =0.0764\end{array}$

Hence, the probability that more than $40\%$ of the people in the sample have high blood pressure.

is $0.0764$ .

To determine

The probability of more than $40\%$ of the people in the sample of $80$ adults having high blood pressure.

Answer to Problem 22E

The probability that more than $40\%$ of the people in the sample of $80$ adults have high blood pressure is $0.0228$ .

Explanation of Solution

Given Information:

  $\begin{array}{l}\stackrel{\wedge }{p}=0.4\\ n=80\\ p=0.3\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

Calculation:

Given that the sample size is $n=80$ and population proportion is $p=0.4$ .

Since, $np=\left(80\right)\left(0.4\right)=32\ge 10$

And $n\left(1-p\right)=\left(80\right)\left(1-0.4\right)=48\ge 10$ .

This means normal approximation can be used to find the probability that more than $40\%$ of the people in the sample have high blood pressure.

Now mean is ${\mu }_p=p=0.3$

The standard deviation is,

  $\begin{array}{c}{\sigma }_p=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.3\left(1-0.3\right)}{80}}\\ =0.05\end{array}$

Calculating the $z$ score for $0.40$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ =\frac{0.40-0.3}{0.05}\\ =2\end{array}$

The required probability will be given as,

  $\begin{array}{c}P\left(\widehat{p}>0.40\right)\\ =1-P\left(\widehat{p}<0.40\right)\\ =1-P\left(z<2\right)\\ =1-0.9772\\ =0.0228\end{array}$

Hence, the probability that more than $40\%$ of the people in the sample of $80$ adults have high blood pressure is $0.0228$ .

To determine

The probability that the proportion of individuals in the sample of $80$ who have high blood pressure is between $0.20\text{ and 0}\text{.35}$ .

Answer to Problem 22E

The probability that the proportion of individuals in the sample of $80$ who have high blood pressure is between $0.20\text{ and 0}\text{.35}$ would be $0.8185$ .

Explanation of Solution

Given Information:

  $\stackrel{\wedge }{p_1}=0.4,\stackrel{\wedge }{p_2}=0.4,n=80,{\mu }_p=p=0.3,{\sigma }_p=0.05$

Calculation:

Calculating the $z$ score for $0.20$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ =\frac{0.20-0.3}{0.05}\\ =-2\end{array}$

Calculating the $z$ score for $0.35$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ =\frac{0.35-0.3}{0.05}\\ =1\end{array}$

The required probability is

  $\begin{array}{c}P\left(0.20<\widehat{p}<0.35\right)\\ =P\left(\widehat{p}<0.35\right)-P\left(\widehat{p}<0.20\right)\\ =P\left(z<1\right)-P\left(z<-2\right)\\ =0.8413-0.0228\\ =0.8185\end{array}$

Hence, the probability that the proportion of individuals in the sample of $80$ who have high blood pressure is between $0.20\text{ and 0}\text{.35}$ would be $0.8185$ .

To determine

The probability that less than $25\%$ of the people in the sample of $80$ have high blood pressure.

Answer to Problem 22E

The required answer is $0.1587$

Explanation of Solution

Given Information:

  $\stackrel{\wedge }{p_1}=0.25,n=80,{\mu }_p=p=0.3,{\sigma }_p=0.05$

Calculation:

Calculating the $z$ score for $0.25$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_P}\\ =\frac{0.25-0.3}{0.05}\\ =-1\end{array}$

The required probability is

  $\begin{array}{c}P\left(\widehat{p}<0.25\right)\\ =P\left(z<-1\right)\\ =0.1587\end{array}$

Hence, the probability that less than $25\%$ of the people in the sample of $80$ have high blood pressure would be $0.1587$ .

To determine

If it is unusual for more than $45\%$ of the individuals in the sample of $80$ to have high blood pressure.

Answer to Problem 22E

Yes, it would be unusual if more than $45\%$ of the individuals in the sample of $80$ had a high blood pressure.

Explanation of Solution

Given Information:

  $\stackrel{\wedge }{p}=0.45,{\mu }_p=p=0.3,{\sigma }_p=0.05$

Calculation:

Calculating the $z$ score for $0.25$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_P}\\ =\frac{0.45-0.3}{0.05}\\ =3\end{array}$

The required probability is.

  $\begin{array}{c}P\left(\widehat{p}<0.45\right)\\ =1-P\left(z<3\right)\\ =1-0.9987\\ =0.0013\end{array}$

The above probability is very low. Hence, it would be unusual if more than $45\%$ of the individuals in the sample of $80$ had a high blood pressure.