Question

In: Statistics and Probability

The systolic blood pressure of adults in the USA is nearly normally distributed with a mean...

The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 119 millimeters of mercury (mmHg) and standard deviation of 25.

Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher. Stage 1 high BP is specified as systolic BP between 140 and 160.

Give your answers rounded to 4 decimal places.

a. What is the probability that an adult in the USA has stage 2 high blood pressure?


b. What is the probability that an adult in the USA has stage 1 high blood pressure?


c. Your doctor tells you you are in the 30th percentile for blood pressure among US adults. What is your systolic BP?
mmHg


d. What is the systolic blood pressure that cuts off the top 2.5% of adults in the USA?
mmHg


Solutions

Expert Solution

Here' the answer to the question. please write back in case you've doubts.

Normal distribution parameters are :
Mean = 119
Stdev = 25
standardizing formula: Z = (X-Mean)/Stdev


a.

P(has stage 2 high blood pressure)
= P(X>160)
= P(Z> (160-119)/25)
= P(Z>1.64)
= 0.0505

b. P(has stage 1 high blood pressure)
= P(140<X<160)
Standardizing,
= P( (140-119)/25 < Z < (160-119)/25)
= P(0.84<Z<1.64)
= 0.94950- 0.8000
= 0.1500

c. The Z for a cumulative of 30 percnetage is -0.5244
The score will be lets "c" such that
P(X<c) = .3
(c-119)/25 = -0.5244
c = 105.8900 mmHg

So, if doctor says you are in 30th percentile for bp amoung US adults you have a systolic BP of 105.8900 mmHg

d.

To be in the top 2.5% of bp you should have a standardized score of at least +1.96
Let such a score be c

So, P(X>c) = .025
or P(X<=c) = 1-.025 = .975
(c-119)/25 = +1.96
c = 168

So, you must have a 168 mmHg systolic BP to be in top 2.5% of the US adults



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