STAT20028
Assignment 2, Term 2 2008
Question 1
5 Marks
a) The histogram below shows a sample of annual percentage returns on investment
portfolios chosen by 50 investment managers from Queensland. Note that the class
interval 0 to less than 6 is represented by the class mark 3, and so on.
Histogram of class mark of annual return
25
20
y15
c
n
e
u
q10
e
r
F
5
0
(i) For this sample, what is the mean annual return?
3
9
15
Answer: Mean = ΣX/n = 15
21
27
Class mark of annual return
(ii) What is the standard deviation for this sample?
Answer: Standard deviation, S =
1 mark
( X  X )
2
1 mark
/(n  1) = 6.6394
Class Mark
Frequency
C
f
Cxf
(C-Xbar)^2*f
3
5
15
720
9
10
90
360
15
20
300
0
21
10
210
360
27
5
135
720
Total:
50
750
2160
Mean:
15
St. deviation:
6.6394
b) Assume that David has contracted to use a computer dating service. He knows that his
personality, background, interest, and so on should make him initially compatible with
30% of the general population. If the computer dating service is doing nothing more than
making random assignments and its clientele is representative of the general population,
what is the probability that out of 10 independent uses of the dating service that David
would be matched with at least two persons with whom he would be initially compatible?
1.5 marks
Answer: p = 0.30, n = 10
P(at least 2)  1  P(0)  P(1)
 1  10C0 (0.3)0 (0.7)0  10C1 (0.3)1 (0.7)9
 1  0.710  10(0.3)(0.7)9
 1  0.0282475  0.1210608
 0.8507
c) The defects in the rubber covering on a particular type of telephone cable follow a
Poisson distribution with a mean rate of 0.00065 per lineal metre. What is the probability
of fewer than 3 defects in a spool of 10,000 lineal metres?
1.5 marks
Answer: Mean, λ = 0.00065×10,000 = 6.5
Fewer than 3 = P(0) + P(1) + P(2)
e6.5 (6.5)0 e6.5 (6.5)1 e6.5 (6.5) 2


0!
1!
2!
 0.001503  0.009772  0.031760

 0.043
Question 2
(a)
5 Marks
A study shows that employees who begin their work day at 8:00am vary their times of
arrival uniformly over the range 7:30am to 8:15am. What is the probability that a random
employee arrives between 8:00am and 8:10am?
1 mark
Answer: Probability = (8:10 – 8:00)/(8:15 – 7:30) = 10/45 = 0.2222
(b)
The errors made in filling a bag of rice are assumed to be normally (Gaussian) distributed
with a mean of zero and a standard deviation of 20 grams. What proportion of bags will
have errors within 30 grams from the desired weight?
1 mark
Answer: 86.64% of bags will have errors within ±30. Z1 = - 30/20 = - 1.5, Z2 = 30/20 = 1.5
From standard normal table , P(Z1<X<Z2) = 0.9332 – 0.0.0668 = 0.8664
(c)
An automatic soft drink dispensing machine is supposed to fill 350ml cups with 325ml of
soft drink. Due to machine tolerance, the amount actually filled is normally distributed
with a variance of 400 ml2. What proportion of cups will overflow?
1 mark
Answer: 10.56% cups will overflow
350  325
 1.25
400
P( Z  1.25)  1  P( Z  1.25)  1  0.8944  0.1056
Z1 
(d)
The CPI for major Australian cities for March quarter 2008 is given below. Can the data be
considered to be normally distributed? (Source: ABS)
1 mark
CPI, Index number for major Australian cities
Mar Qtr 2008
Sydney
161.7
Melbourne
160.6
Brisbane
165.6
Adelaide
165.5
Perth
162.5
Hobart
161.3
Darwin
158.5
Canberra
163.0
Answer: The normal probability plot given below and obtained by PHStat is not really a straight line.
Therefore, it is not normally distributed in a strict sense, but roughly it can be considered as normal
distribution. However, more observations are needed for a conclusive decision.
(e)
Given an arrival process follows an exponential distribution that has a mean of 20, what is
the probability that the arrival time will be less than 2?
1 mark
Answer: P(Arrival time < 2)  1  e  (2)  1  e40  1  4.248 1018 = 1.00
Question 3
(a)
5 Marks
The population of a variable is known to have Gamma distribution. We wish to make some
inferences about the variable using the theories of normal distribution. How can we achieve
that through sampling?
1 mark
Answer: Gamma distribution is a skewed distribution. To apply normal distribution theory the
Central Limit Theorem should take effect, and for that purpose, we need to select the sample size
sufficiently large. Specifically, the sample size should be greater than 30.
(b)
The price-earnings ratios of a very large (essentially infinite) set of common stocks are
approximately normally distributed with a mean of 15 and a standard deviation of 5. If a
large number of samples, each of size 25, are selected from the population of stocks and a
sample mean is computed for each sample, what will be the expected value and the variance
of the large number of sample means for the price-earnings ratios?
2 marks
Answer: The expected value of sample means = 15
The variance of the sample means = (5/√25)2 = 1
(c)
Fifty-five percent of Rocky Fitness Centre customers are women. Many samples with n =
25 were taken from this Centre’s customer list. What is the standard deviation of the
proportion of women customers found in these samples?
2 marks
Answer: standard deviation =
P(1  P)
0.55(1  0.55)

 0.099
n
25
Question 4
(a)
5 Marks
A health service provider has been reimbursed by Medibank Private for a large number of
services. It appears that overpayments may have occurred on many of the reimbursements
due to errors made during the billing process or owing to upcoding (billing for a more
complicated service than was actually provided). Find the 95% confidence interval estimate
of the mean overpayment per service given the following random sample of overpayments.
$110, $130, $90, $100, $120, $105, $95, $115, $125
2 marks
Answer: Sample size is small and sample probability distribution is not known, therefore, tdistribution should be used. Sample mean = 110, standard deviation = 13.693
Confidence interval = X  tn 1,0.025
(b)
S
13.693
= (99.47, 120.53)
 110  2.306
n
9
A local government agency has just purchased a new computer software product. The
agency chief is interested in determining how many of the local government’s 3,900
employees have previous experience with similar software product. She takes a sample of
50 people and finds 6 who have had this experience. Construct a 95% confidence interval
for the proportion of people who have had this experience.
2 marks
Answer: p = 6/50 = 0.12
Confidence interval  p  Z
(c)
p(1  p)
0.12(1  0.12)
= (0.03, 0.21)
 0.12  1.96
n
50
BP wishes to estimate the mean amount of water that has seeped into the fuel storage tanks
at its refineries in Brisbane. A preliminary sample of n = 21 tanks showed that the standard
deviation, s = 45 litres. How much larger should the sample be in order to estimate the
mean water content of the tanks to within ±10 litres with 95% confidence?
1 mark
Answer: Sampling error, e = 10, s = 45, we do not know t, therefore start with the Z value which
is 1.96. Therefore, n 
Z 2 s 2 (1.96)2 (45)2

 77.79 ≈ 78. (Since the required sample size is
e2
(10)2
greater than 30 we do not have to refine the result with t-value instead of the Z-value).
Thus, the sample size has to be increased by 57 (=78-21) additional samples.