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

0.0224

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>(140-114.8)/(13.1) =P(Z>1.923)\\`

=0.5-0.4726

=0.0274

Answer 2

0.0274

Explanation:

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

Calculate

`Z=(X-\mu)/(\sigma)`

for
`X=140:`

`Z=(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.`