Question

The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 122 and standard deviation of 16 . Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher.a. Around what percentage of adults in the USA have stage 2 high blood pressure? Give your answer rounded to two decimal places.b. If you sampled 2000 people, how many would you expect to have BP> 160? Give your answer to the nearest person. Please explain how to find this in the calculator step by step.Note: I had a bit of an issue encoding rounded answers, so try rounding both up and down if there's an issue!c. Stage 1 high BP is specified as systolic BP between 140 and 160. What percentage of adults in the US qualify for stage 1?d. Your doctor tells you that you are in the 30th percentile for blood pressure among US adults. What is your systolic BP? Round to 2 decimal places.

Answer 1

Answer:

A) 0.0087

B) 17

C) 0.1205

D) 113.68

Step-by-step explanation:

Given data

Mean = 122

Standard deviation = 16

(a)

We need to calculate $P(\ bar X \geq 160) = 1 - P( \bar X <160)$

$Z = \frac{(160-122)}{16} = 2.375$

from Z table we found p-value

$P(X \bar \geq 160) = 1- 0.9913 = 0.0087$

(b)

If sample size = 2000

so there are$(2000\times 0.0087) = 17.4 or 17$ people who have bp > 160

(c)

$P(140 < \bar X < 160) = P(\bar X< 160) - P( \bar X < 140)$

$Z = \frac{(160-122)}{16} = 2.375$

$Z = \frac{(140-122)}{16} = 1.125$

So from Z table we found p-value

$P(140 < \bar X < 160) = 0.9913 - 0.8708 = 0.1205$

(d)

P-value = 0.3

from Z table,  Z-score is -0.52

So $-0.52 = \frac{(\bar X -122)}{16}$

$-8.32 = \bar X -122$

$\bar X = 113.68$