Question

For women aged 18-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1 (based on data from the National Health Survey). If 15 women in that age bracket are randomly selected, find the probability that their mean systolic blood pressure is between 110 and 115.

Select one:

a. 41.89%

b. 39.60%

c. 49.70%

d. None of other answers is neccessary true.

e. 44.56%

Answer 1

Solution:

We are given

µ = 114.8

σ = 13.1

n = 15

We have to find P(110<Xbar<115)

P(110<Xbar<115) = P(Xbar<115) - P(Xbar<110)

First find P(Xbar<115)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (115 - 114.8)/[13.1/sqrt(15)]

Z = 0.05913

P(Z< 0.05913) = P(Xbar<115) = 0.523576

(by using z-table)

Now find P(Xbar<110)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (110 - 114.8)/[13.1/sqrt(15)]

Z = -1.41911

P(Z< -1.41911) = P(Xbar<110) = 0.077934

(by using z-table)

P(110<Xbar<115) = P(Xbar<115) - P(Xbar<110)

P(110<Xbar<115) = 0.523576 - 0.077934

P(110<Xbar<115) = 0.445642

Required probability = 0.445642

Answer: e. 44.56%