Question

1. I. Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 6.8-in and a standard deviation of 0.8-in. Due to financial constraints, the helmets will be designed to fit all men except those with head breadths that are in the smallest 2.6% or largest 2.6%.
What is the minimum head breadth that will fit the clientele?
min = inches
What is the maximum head breadth that will fit the clientele?
min = inches
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

II. The systolic blood pressure of adults in a large city is nearly normally distributed with a mean of 117 and standard deviation of 23 .
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.
people
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 you are in the 30th percentile for blood pressure among US adults. What is your systolic BP? Round to 2 decimal places.
lbs

Answer 1

Let X denote the head breadth.

a)P(X<x)=0.026 which implies x=5.2 inch

b)P(X>x)=0.026 which iplies x=8.4 inch

Let Y denote systolic blood pressure.

a)percentage of people have stage 2 systolic BP=100*0.03077211=3.08% approximately

b)number of people havinh=g stage 2 BP among 2000 peolple=2000*0.03077211=62 approximately

c)percentage of adults have stage 1 systolic BP=100*0.1278831=12.79% approximately

d)30th percentile of the distribution=104.94

R CODE:

> 6.8+0.8*qnorm(.026)
[1] 5.245493
> 6.8+0.8*qnorm(.026,lower.tail=FALSE)
[1] 8.354507
>
>
> pnorm(160,117,23,lower.tail=FALSE)
[1] 0.03077211

> 2000*0.03077211
[1] 61.54422
> pnorm(160,117,23,lower.tail=TRUE)-pnorm(140,117,23,lower.tail=TRUE)
[1] 0.1278831
> qnorm(0.30,117,23)
[1] 104.9388