Question

2. The ages of subscribers to a certain newspaper are normally distributed with mean 37.5 years and standard deviation of 5.1. What is the probability that the age of a random subscriber is (a) more than 37.5 years

(b) between 30 and 40 years?

Answer 1

Solution :

Given that ,

mean = = 37.5

standard deviation = = 5.1

(a)

P(x > 37.5) = 1 - P(x < 37.5)

= 1 - P[(x - ) / < (37.5 - 37.5) / 5.1)

= 1 - P(z < 0)

= 1 - 0.5

= 0.5

Probability = 0.5

(b)

P(30 < x < 40) = P[(30 - 37.5)/ 5.1) < (x - ) /  < (40 - 37.5) / 5.1) ]

= P(-1.47 < z < 0.49)

= P(z < 0.49) - P(z < -1.47)

= 0.6879 - 0.0708

= 0.6171

Probability = 0.6171