Question

Assume that the mean systolic blood pressure of normal adults is 120 milliliters of mercury (mmHg) and the standard deviation is 5.6. Assume the variable is normally distributed. a) If an individual is selected, find the probability that the individual’s pressure will be between 118 and 122.8 mmHg. b) If a sample of 40 adults is randomly selected, find the probability that the sample mean will be more than 121 mmHg.

Answer 1

Solution :

Given that ,

mean = = 120

standard deviation = = 5.6

a) P(118 < x < 122.8) = P[(118 - 120) / 5.6) < (x - ) /  < (122.8 - 120) / 5.6) ]

= P( - 0.36 < z < 0.5)

= P(z < 0.5) - P(z < - 0.36)

Using z table,

= 0.6915 - 0.3594

= 0.3321

b) = 40

= / n = 5.6 / 40 = 0.89

P( > 121 ) = 1 - P( < 121)

= 1 - P[( - ) / < (121 - 120) / 0.89]

= 1 - P(z < 1.12)   

= 1 - 0.8686

= 0.1314