Question

Problem 1: The average Saturday attendance at a movie theater is 974 people with a standard deviation of 54 people.

Part A: What is the probability that less than 900 people will attend this coming Saturday?

Part B: What is the probability of between 875 and 1075 people will attend this Saturday?

Part C: Eighty percent of Saturday attendances will be less than how many people?

Part D: The movie theater manager wants to determine a staffing level such that 98% of the time she can service the customers. How many customers should she set a staffing plan to serve?

Answer 1

Solution :

Given that ,

mean = = 974

standard deviation = = 54

Part A:

P(x < 900) = P[(x - ) / < (900 - 974) / 54]

= P(z < -1.37)

= 0.0853

Probability = 0.0853

Part B:

P(875 < x < 1075) = P[(875 - 974)/ 54) < (x - ) /  < (1075 - 974) / 54) ]

= P(-1.83 < z < 1.87)

= P(z < 1.87) - P(z < -1.83)

= 0.9693 - 0.0336

= 0.9357

Probability = 0.9357

Part C:

Using standard normal table,

P(Z < z) = 80%

P(Z < 0.84) = 0.8

z = 0.84

Using z-score formula,

x = z * +

x = 0.84 * 54 + 974 = 1019.36

people = 1019.39

Part D:

Using standard normal table,

P(Z < z) = 98%

P(Z < 2.05) = 0.98

z = 2.05

Using z-score formula,

x = z * +

x = 2.05 * 54 + 974 = 1084.7

1084.7 customers should she set a staffing plan to serve