Probability question

Question

The systolic blood pressure of adults in the USA is nearly normally distributed with a mean of 119 and standard deviation of 18.

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?

b. If you sampled 500 people, how many would you expect to have BP> 160? 

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? 

Answer 1

a) if x is systolic blood pressure, then want to find probability that x is >= 160.

z = (x - mean)/standard deviation = (x - 119)/18 is normally distributed with mean 0 and standard deviation 1

Transform x to z:

P(x > 160) = P(z > (160-119)/18) = P(z > 2.78) = P(z < -2.78) = 0.0027 (from standard normal distribution table, which 0.27%.

b) Since probability of BP > 160 is 0.027. Then out of 500 people, then 500 * 0.027 = 13.5 people would have BP > 160. (or between 13 and 14 people)>

c) Now we need to find P(140 < x < 160).

Transform to z:

P( (140-119)/18 < z < (160-119/18))

= P( 1.17 < z < 2.28) = P(z < 2.28) - P(z<1.17) = 0.9887 - 0.8790 = 0.1097, so 10.97%