Chapter 7.2, Problem 27E

Check your blood pressure: In a recent study, the Centers for Disease Control and Prevention repotted that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.5 and standard deviation 9.9.

1. What proportion of women have blood pressures lower than 70?
2. What proportion of women have blood pressures between 75 and 90?
3. A diastolic blood pressure greater than 90 is classified as hypertension (high blood pressure). What proportion of women have hypertension?
4. Is it unusual for a woman to have a blood pressure lower than 65?

To determine

The proportion of women having blood pressure lower than $70$ .

Answer to Problem 27E

  $14.46\%$ of adult women have blood pressure lower than $70$ .

Explanation of Solution

Given information:

In a study it was found that the diastolic blood pressure of adult women in US is normally distributed with mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

Formula used:

Formula for probability for normal distribution

  $P(X=x)=P\left(\frac{X-\mu }{\sigma }=\frac{x-\mu }{\sigma }\right)=P\left(Z=\frac{x-\mu }{\sigma }\right)$

where $Z\text{ is z-score, }\mu \text{ is mean and }\sigma \text{ is standard deviation}\text{.}$

Calculation:

Here mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

The proportion of population which is less than $70$ is calculated as-

  $\begin{array}{l}P(X<70)=P\left(\frac{x-\mu }{\sigma }<\frac{70-80.5}{9.9}\right)\\ \Rightarrow P(X<70)=P\left(Z<-1.06\right)\\ \text{Using standard normal distribution table,}\\ P(X<70)=0.14457\\ \Rightarrow P(X<70)=14.46\%\end{array}$

Hence, $14.46\%$ of adult women have blood pressure lower than $70$ .

To determine

The proportion of women having blood pressure between $75\text{ and 90}$ .

Answer to Problem 27E

  $54.37\%$ of adult women have blood pressure between $75\text{ and 90}$ .

Explanation of Solution

Given information:

In a study it was found that the diastolic blood pressure of adult women in US is normally distributed with mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

Formula used:

Formula for probability for normal distribution

  $P(X=x)=P\left(\frac{X-\mu }{\sigma }=\frac{x-\mu }{\sigma }\right)=P\left(Z=\frac{x-\mu }{\sigma }\right)$

where $Z\text{ is z-score, }\mu \text{ is mean and }\sigma \text{ is standard deviation}\text{.}$

Calculation:

Here mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

The proportion of population which is between $75\text{ and 90}$ is calculated as-

  $\begin{array}{l}P(75<X<90)=P(X<90)-P(X<75)\\ \Rightarrow P(75<X<90)=P\left(\frac{x-\mu }{\sigma }<\frac{90-80.5}{9.9}\right)-P\left(\frac{x-\mu }{\sigma }<\frac{75-80.5}{9.9}\right)\\ \Rightarrow P(75<X<90)=P(Z<0.96)-P(Z<-0.56)\\ \text{Using standard normal distribution table,}\\ P(75<X<90)=0.83147-0.28774\\ \Rightarrow P(75<X<90)=0.54373\\ \Rightarrow P(75<X<90)=54.37\%\end{array}$

Hence, $54.37\%$ of adult women have blood pressure between $75\text{ and 90}$ .

To determine

The proportion of women who has hypertension i.e. having blood pressure higher than $90$ .

Answer to Problem 27E

  $16.85\%$ of adult women have hypertension.

Explanation of Solution

Given information:

In a study it was found that the diastolic blood pressure of adult women in US is normally distributed with mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

A blood pressure higher than $90$ is termed as hypertension.

Formula used:

Formula for probability for normal distribution

  $P(X=x)=P\left(\frac{X-\mu }{\sigma }=\frac{x-\mu }{\sigma }\right)=P\left(Z=\frac{x-\mu }{\sigma }\right)$

where $Z\text{ is z-score, }\mu \text{ is mean and }\sigma \text{ is standard deviation}\text{.}$

Calculation:

Here mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

The proportion of population which is greater than $90$ is calculated as-

  $\begin{array}{l}P(X>90)=1-P(X\le 90)\\ \Rightarrow P(X>90)=1-P\left(\frac{x-\mu }{\sigma }\le \frac{90-80.5}{9.9}\right)\\ \Rightarrow P(X>90)=1-P\left(Z\le 0.96\right)\\ \text{Using standard normal distribution table,}\\ P(X>90)=1-0.83147\\ \Rightarrow P(X>90)=0.16853\\ \Rightarrow P(X>90)=16.85\%\end{array}$

Therefore, $16.85\%$ of women have hypertension.

To determine

If it is unusual for a women to have blood pressure lower than $65$ .

Answer to Problem 27E

Women having blood pressure less than $65$ is a usual value.

Explanation of Solution

Given information:

In a study it was found that the diastolic blood pressure of adult women in US is normally distributed with mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

Calculation:

Here mean $\mu =80.5$ and standard deviation $\sigma =9.9$ .

If given value is in between the $\pm 2\sigma$ point value to the mean ( $2\sigma$ below and $2\sigma$ above the mean) called as usual value otherwise that is unusual value.

  $\begin{array}{l}\mu -2\sigma <x<\mu +2\sigma \\ \Rightarrow 80.5-2\times 9.9<x<80.5+2\times 9.9\\ \Rightarrow 60.7<x<100.3\end{array}$

Here women having blood pressure less than $65$ is in between the values $60.7\text{ and }100.3$ .

Therefore, women having blood pressure less than $65$ is a usual value.