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 $1370 should be taken for a.
1 / 10
2 / 10
3 / 10
4 / 10
5 / 10
6 / 10
7 / 10
8 / 10
9 / 10
10 / 10
P /100
2) The probability that a drug cures a patient is 90%. Use a normal distribution to approximate the probability that at least 56 out of 60 people who take the drug are cured.
Answer. Since nθ = 60 ×0.90 = 54 ≥ 5 and n(1−θ) = 60×0.10 = 6 ≥ 5, the Binomial Distribution can be considered normal with µ = nθ = 60 × 0.90 = 54 and σ = pnθ(1 − θ) =√
60 × 0.90 × 0.10 = 3√515. Thus
P (X ≥ 56) = P (X > 55.5) ≈ P Z > 55.5 − 54
3√ 15 5
!
= P Z >
√15 6
!
= P (Z > 0.65)
≈ 0.5 − 0.2422 = 0.2578.
3) A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 120 homeowners selected from this area showed that they pay an average of 1575$ 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.
Answer. Since
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?
Answer.
35 ≥ 2.17215
√n ⇒√
n ≥ 2.17 × 215
35 ⇒ n ≥ 2.17 × 215 35
2
= 177.69 nmin= 178.
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.
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× e−21
10! = 0.003485.
