Question

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6. What proportion of women have blood pressures between 88.1 and 89.4

Answer 1

Answer:

Proportion of women having blood pressures between 88.1 and 89.4 is 3.99% or close to 4%.

Step-by-step explanation:

We are given that a recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6.

Let X = blood pressures of adult women in the United States

So, X ~ N($\mu=80.3,\sigma^{2} =8.6^{2}$)

The z score probability distribution is given by;

              Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)  

where, $\mu$ = population mean

           $\sigma$ = population standard deviation

So, Probability that women have blood pressures between 88.1 and 89.4 is given by = P(88.1 < X < 89.4) = P(X < 89.4) - P(X $\leq$ 88.1)

   P(X < 89.4) = P( $\frac{X-\mu}{\sigma}$ < $\frac{89.4-80.3}{8.6}$ ) = P(Z < 1.05) = 0.85314

   P(X $\leq$ 88.1) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{88.1-80.3}{8.6}$ ) = P(Z $\leq$ 0.89) = 0.81327

Therefore, P(88.1 < X < 89.4) = 0.85314 - 0.81327 = 0.0399 or approx 4%

Hence, proportion of women have blood pressures between 88.1 and 89.4 is 4%.