Question

For a woman age 18 to 24, systolic blood pressure’s (in mm of Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1 (based on the data from the national health survey).

A. If a woman between the ages of 18 and 24 is randomly selected, find the probability that her systolic blood pressure is above 120.

B. If 30 women in that age bracket are randomly selected, find the probability that the main systolic blood pressure is greater than 120.

Answer 1

Solution :

Given that ,

mean = = 114.8

standard deviation = = 13.1

A. P(x > 120)

= 1 - P( x < 120)

= 1 - P[(x - ) / < (120 - 114.8) / 13.1]

= 1 - P(z < 0.40)

Using z table,

= 1 - 0.6554

= 0.3446

Probability = 0.3446

B. n = 30

= 114.8

= / n = 13.1 / 30 = 2.3917

P( > 120)

= 1 - P( < 120)

= 1 - P[( - ) / < (120 - 114.8) / 2.3917]

= 1 - P(z < 2.17)

Using z table,    

= 1 - 0.9850

= 0.0150

Probability = 0.0150