# Let X be the lifetime of the item

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MCB1007 Introduction to Probability and Statistics Fall 2017-2018

Final Exam

December 18, 2017

No:

Name:

Section:

Justify your answers to get full credit − Exam Duration 90 Minutes 1) An expensive item is being insured against early failure. The lifetime of the item

is normally distributed with an expected value of 7 years and a standard deviation of 2 years. The insurance will pay a dollars if the item fails during the first or second year and a/2 dollars if the item fails during the third or fourth year. If a failure occurs after the fourth year, then the insurance pays nothing. How to choose a such that the expected value of the payment per insurance is \$50?

Answer. Let X be the lifetime of the item. The expected value of the payment per insurance is

aP (X ≤ 2) + 1

2aP (2 < X ≤ 4) = aP



Z ≤ 2 − 7 2

 +1

2aP 2 − 7

2 ≤ Z < 4 − 7 2



= aP (Z ≤ −2.5) + 1

2aP (−2.5 ≤ Z < −1.5)

= a(0.5 − 0.4939) + 1

2a(0.4938 − 0.4332)

= a(0.0062 + 0.0303) = 0.0365a.

Thus

0.0365a = 50 ⇒ a = 1369.863014.

Hence the value \$1370 should be taken for a.

1 / 10

2 / 10

3 / 10

4 / 10

5 / 10

6 / 10

7 / 10

8 / 10

9 / 10

10 / 10

P /100

2) The probability that a drug cures a patient is 90%. Use a normal distribution to approximate the probability that at least 56 out of 60 people who take the drug are cured.

Answer. Since nθ = 60 ×0.90 = 54 ≥ 5 and n(1−θ) = 60×0.10 = 6 ≥ 5, the Binomial Distribution can be considered normal with µ = nθ = 60 × 0.90 = 54 and σ = pnθ(1 − θ) =√

60 × 0.90 × 0.10 = 3515. Thus

P (X ≥ 56) = P (X > 55.5) ≈ P Z > 55.5 − 54

3 15 5

!

= P Z >

√15 6

!

= P (Z > 0.65)

≈ 0.5 − 0.2422 = 0.2578.

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3) A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of 1575\$ per month. The population standard deviation of such mortgages is 215\$.

a) Find a %97 confidence interval for the mean amount of mortgage paid per month by all homeowners in this area.

Answer. For zα/2 = 2.17, σ = 215, n = 120, ¯x = 1575 we have

x − z¯ α/2 σ

√n < µ < ¯x + zα/2

√σ n

1575 − 2.17 215

√120 < µ < 1575 + 2.17 215

√120 1532.41 < µ < 1617.59.

b) With what degree of confidence could we say that the mean amount of mortgage paid per month is 1575 ± 34.

34 = zα/2 215

√120 ⇒ zα/2= 34√ 120

215 = 1.73 the degree of confidence is

2 × 0.4582 = 0.9164.

c) What should be the minimum sample size, if error does not exceed 35?

35 ≥ 2.17215

√n ⇒√

n ≥ 2.17 × 215

35 ⇒ n ≥ 2.17 × 215 35

2

= 177.69 nmin= 178.

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4) Number of customers who visited a company that is set up 13 days ago are recorded in the following table. Estimate mean number, median and the modal class of customers.

Number of Days Number of Customers

2 − 4 14

5 − 7 13

8 − 10 3

11 − 13 20

Answer. Mean: ¯x = 14 × 3 + 13 × 6 + 3 × 9 + 20 × 12

50 = 7.74.

Median: 4.5 + 7.5 − 4.5

13 × 11.5 = 7.15.

Modal Class: 11 − 13.

5) From the set of numbers {1, 5, 9} a random sample of size 2 will be selected without replacement.

a) Find the mean and the standard deviation of the population Answer. f (x) = 1

3, x = 1, 5, 9.

µX = E(X) =P

xxf (x) = 1 + 5 + 9 3 = 5, E(X2) =P

xx2f (x) = 12+ 52+ 92

3 = 107

3 , σX =pE(X2) − (E(X))2 =r 107

3 − 52 = 4√ 6

3 = 3.27.

b) Find the mean and the standard deviation of the sampling distribution Answer.

(x1, x2) (1, 5) (1, 9) (5, 9)

¯

x 3 5 7

¯

x 3 5 7

f (¯x) 1/3 1/3 1/3 µx¯ = µx= 5,

σx¯ = σX

√n

r N − n N − 1 =

4 6

3

2

r 3 − 2

3 − 1 = 2√ 6

3 = 1.63.

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6) There are 300 employees in a company, 240 of whom are trade union member. 6 employees will be selected randomly among all employees for the commission, what is the probability that 4 of which are trade union member?

Answer. Hypergeometric distribution with parameters N = 300, M = 240, n = 6 and x = 4:

P (X = 4) =

240 4

 60 2



300 6

 = 0.2478.

7) A dry cleaner receives 3 complaints on average in a day. What is the probability that this dry cleaner receives 10 complaints in a week.

Answer. Poisson distribution with parameters λ = 3 × 7 = 21 and x = 10:

P (X = 10) = 2110× e21

10! = 0.003485.

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normally distributed with a mean of 7 years and a standard deviation of 2 years. The insurance will pay a dollars if the item fails during the first or second year and a/2 dollars