Question

A woman is randomly selected from the 18-24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. What is the probability this woman has a systolic blood pressure greater than 140?

Answer 1

Answer:

0.0224

Step-by-step explanation:

Given that systolic blood pressures of women are normally distributed.

IF X is the systolic bp of women X is N(114.8, 13.1)

Required probability

=  the probability this woman has a systolic blood pressure greater than 140

=$P(X>140)$

=$P(Z>\frac{140-114.8}{13.1} =P(Z>1.923)$

=0.5-0.4726

=0.0274

Answer 2

Answer:

0.0274

Step-by-step explanation:

The mean is $\mu =114.8$ and the standard deviation is $\sigma =13.1.$

Calculate

$Z=\dfrac{X-\mu}{\sigma}$

for $X=140:$

$Z=\dfrac{140-114.8}{13.1}\approx 1.9237.$

If $X\sim N(114.8,\ 13.1),$ then $Z\sim N(0,1)$

and

$Pr(X>140)=Pr(Z>1.9237).$

Use table for normal distribution probabilities to get that

$Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.$