Question

- 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**)

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

minutes?

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

minutes?

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

15 minutes?

- 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**

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