Question

7) If the systolic blood pressure is normally distributed in the population with a mean=120, SD=20, X~N(120, 20) what is the probability of a random person having systolic blood pressure of 130 or greater (hint: calculate the Z score, then find the probability in the right tail)

Answer 1

Explanation:

Given that the systolic blood pressure is normally distributed in the population with a mean=120, SD=20.

the probability of a random person having systolic blood pressure of 130 or greater

Convert 130 to Z score as

`z=(130-120)/(20) =0.5`

Required probability=

P(Z\geq 0.5) =

=0.5-0.1915

=0.3085