Question

4. The length of incoming calls at the call centre of a major telecommunication service provider follows a Normal distribution with an average of 6.5 minutes. If the standard deviation of the distribution is 4 minutes, answer the following questions: (8 points)

1. What is the probability that the length of an incoming call is longer than 8.5

minutes?

2. What is the probability that the length of an incoming call is shorter than 5

minutes?

3. What is the probability that the length of an incoming call is between 5.5 to

15 minutes?

4. 25% of the incoming calls is longer than how many minutes?

Answer 1

Norml distribution: P(X < A) = P(Z < (A-mean)/standard deviation)

Mean = 6.5 minutes

Standard deviation = 4

a. P(X > 8.5) = 1 - P(X < 8.5)

= 1 - P(Z < (8.5 - 6.5)/4)

= 1 - P(Z < 0.5)

= 1 - 0.6915

= 0.3085

b. P(X < 5) = P(Z < (5 - 6.5)/4)

= P(Z < -0.375)

= 0.3538

c. P(5.5 < X < 15) = P(X < 15) - P(X < 5.5)

= P(Z < (15 - 6.5)/4) - P(Z < (5.5 - 6.5)/4)

= P(Z < 2.125) - P(Z < -0.25)

= 0.9832 - 0.4013

= 0.5819

d. Let 25% of calls be longer than M miniutes

P(X > M) = 0.25

P(X < M) = 1 - 0.25 = 0.75

P(Z < (M - 6.5)/4) = 0.67

M = 9.18