Question

Question 7
A telephonist claims that he gets 10 calls every five minutes. To demonstrate this to his boss he makes a tape lasting five minutes. What are the probabilities that he gets:
(a) no calls in the five minutes; (2)
(b) less than three calls; and (5)
(c) exactly 10 calls? (2) [9]

Question 8
Assume that matric marks are standardised to have a mean of 52% and a standard deviation of 16% (and assume that they have a normal distribution). In a class of 100 students estimate how many of them:
(a) pass (in other words get more than 33,3%); (4)
(b) get A's (more than 80%); and (4)
(c) get B's (between 70% and 80%). (6) [14]

Question 9

As manager of a company you know that the distribution of completion times for an assembly operation is a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. If you have to award bonuses to the top 10% of your workers what time would you use as a cut-off time?

Answer 1

7) = 10

It is A poision distribution.

P(X = x) = e/X!

A) P(X = 0) = e^(-10) * (10)^0/0! = 0.000045

B) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

=  e^(-10) * (10)^0/0! +  e^(-10) * (10)^1/01! +  e^(-10) * (10)^2/2! = 0.00277

C) P(X = 10) = e^(-10) * (10)^10/10! = 0.12511

8) a) P(X > 33.3)

= P((X - )/ > (33.3 - )/)

= P(Z > (33.3 - 52)/16)

= P(Z > -1.17)

= 1 - P(Z < -1.17)

= 1 - 0.1210

= 0.8790

Expected number of pass students = 100 * 0.8790 = 87.90 = 88

B)

P(X > 80)

= P((X - )/ > (80 - )/)

= P(Z > (80 - 52)/16)

= P(Z > 1.75)

= 1 - P(Z < 1.75)

= 1 - 0.9599

= 0.0401

Expected number students get A = 100 * 0.0401 = 4.01= 4

C) P(70 < X < 89)

= P((70 -  )/ < (X -  )/ < (80 -  )/)

= P((70 - 52)/16 < Z < (80 - 52)/16)

= P(1.13 < Z < 1.75)

= P(Z < 1.75) - P(Z < 1.13)

= 0.9599 - 0.8708

= 0.0891

Expecting number of students get B = 0.0891 * 100 = 9.91 = 10

9) P(X > X) = 0.10

Or, P((X -  )/ > (x -  )/) = 0.10

Or, P(Z > (x - 120)/20) = 0.10

Or, P(Z < (X - 120)/20) = 0.90

Or, (X - 120)/20 = 1.28

Or, X = 1.28 * 20 + 120

Or, X = 145.6