Question

In a recent study, the Centers for Disease Control reported that diastolic blood pressures of adult women in the U.S. are approximately normally distributed with a mean of 80.5 and standard deviation 9.9. Round all answers to 4 decimal places. a. What proportion of women have blood pressures lower than 70? b. Find the 30th percentile for blood pressures. C. Suppose a sample of 150 women is taken. What is the probability that the sample mean blood pressure would be between 75 and 80?

Answer 1

Here

$mean=\mu =80.5$

Standard deviation = $\sigma$ = 9.9

a)

$P(X<70)=P(\frac{X-\mu }{\sigma }<\frac{70-80.5}{9.9})$

= P(Z < -1.06

= 0.1446 ( from standard normal table )

Therefore, the proportion of women that have blood pressures lower than 70 is 0.1446

b)

P(X <C) = 0.30

$P(\frac{X-\mu }{\sigma }<\frac{x-80.5}{9.9})=0.30$

  $P(Z<z)=0.30$

From Standard normal table, the z-critical value corresponding to the closest probability of 0.30 is -0.52

i.e. z = -0.52

Now

$X=\mu +z\sigma =80.5+(-0.52)*9.9= 75.352$

The 30th percentile for blood pressure is 75.352

c)

Here

n = 150

$P(75<\overline{x}<80)=P(\frac{75-80.5}{9.9/\sqrt{150}}<\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}<\frac{80-80.5}{9.9/\sqrt{150}})$

  $=P(-6.80<Z<-0.62)$

  $=P(Z<-0.62)-P(Z<-6.80)$

  $=0.2676$

The probability that the sample mean blood pressure would be between 75 and 80 is 0.2676