Question

Please provide the solution to both of the parts. Please its urgent.

a) A travel agency acts as a selling agent for two airline companies. The two airline companies are charged the same monthly fee for this service and it is assumed that they will each receive approximately half the calls coming in. Based on this assumption, find the probability that in the next 150 calls 90 or more will be for one agent. (Use normal approximation of the binomial distribution.)                         

b) In 2017, it is found that 60% of video games purchasers are men. If a random sample is taken of 15 people who purchase video games, what is the approximate probability that there will be an exactly 11 men?

Answer 1

a)

The probability of call to an agent, p = 0.5

Total number of calls, n = 150

Use normal approximation of the binomial distribution, the mean and standard deviation of the normal distributuin is,

Mean = np = 150 * 0.5 = 75

Standard deviation = √(np(1-p)) = √(150 * 0.5 * (1-0.5)) = 6.124

Probability that in the next 150 calls 90 or more will be for one agent =

= 1 - Probability that in the next 150 calls, 89 or less will be for one agent

= 1 - P[X 89]

= 1 - P[Z (89 - 75) / 6.124]

= 1 - P[Z 2.286]

= 1 - 0.9889

= 0.0111

b)

Probability that the video games purchasers are men, p = 0.6

Total number of people in sample, n = 15

Use normal approximation of the binomial distribution, the mean and standard deviation of the normal distributuin is,

Mean = np = 15 * 0.6 = 9

Standard deviation = √(np(1-p)) = √(15 * 0.6 * (1-0.6)) = 1.897

Probability that there will be an exactly 11 men = P[X = 11]

= P[Z = (11 - 9) / 1.897]

= P[Z = 1.0543]

= 0.2288