Question

Suppose the average time a visitor spends at an Art Gallery is 35 minutes. Assume the visit lengths are normally distributed with a standard deviation of 11 minutes. If a visitor is selected at random, what is the probability that they spend:
(a) less than 20 minutes at the exhibit?
(b) more than 40 minutes at the exhibit?

Answer 1

Solution :

Given that ,

mean = = 35

standard deviation = = 11

(a)

P(x < 20) = P[(x - ) / < (20 - 35) / 11]

= P(z < -1.36)

= 0.0869

Probability = 0.0869

(b)

P(x > 40) = 1 - P(x < 40)

= 1 - P[(x - ) / < (40 - 35) / 11)

= 1 - P(z < 0.45)

= 1 - 0.6736

= 0.3264

Probability = 0.3264