Assessment item 2—Assignment 2
Due date:
5:00pm, 11 September 2009
Weighting:
20%
ASSESSMENT
2
Length:
Question 1
5 Marks
a) The histogram below shows a sample of annual percentage returns on investment
portfolios chosen by 60 investment managers from Queensland. Note that the class
interval 1 to less than 3 is represented by the class mark 2, and so on.
Histogram of class mark of annual return
25
20
15
F
r
e 10
q
u
e 5
n
c
y 0
2
4
6
8
10
Class mark of annual return
(i) For this sample, what is the expected value of annual return?
Answer: Expected value = mean =
1 mark
2  15  4  10  6  20  8  10  10  5
= 5.3333
15  10  20  10  5
(ii) What is the variance of annual return for this sample?
1 mark
Answer: Variance = E[( X  X ) 2 ] = (2-5.3333)2(15/59) + … + (10-5.3333)2(5/59) = 6.3277
b) A student majoring in accounting at CQ University is trying to decide on the number of
firms to which she should apply. Given her work experience, academic results and
extracurricular activities, she has been told by a placement counsellor that she can expect
to receive a job offer from 70% of the firms to which she applies. Wanting to save time,
the student applied to five firms only. Assuming the counsellor’s estimate is correct, find
the probability that the student receives at least one offer.
1.5 marks
Answer: P(at least one offer) = 1 – P(no offer). It is a case of binomial distribution
n
P(no offer) =   p x (1  p ) n  x =
 x
 5
 (0.7) 0 (0.3) 5 = 0.00243
 0
Therefore, P(at least one offer) = 1 – 0.00243 = 0.99757
c) The marketing manager of a mail-order company has noted that she usually receives 10
complaint calls from customers during a week (consisting of five working days), and that
the calls occur at random. Find the probability of her receiving four such calls on a single
day.
1.5 marks
Answer: It is a case of Poisson distribution, where λ = 2 calls per day and X = 4 calls.
P(X  4) 
e    X e 2 2 4

= 0.0902. Therefore, the probability is 0.0902 of getting 4 calls in
X!
4!
a day.
Question 2
(a)
5 Marks
A hospital receives a pharmaceutical delivery each morning at a time that varies uniformly
between 7:00am and 8:00am. Find the probability that the time of delivery will be within
one standard deviation of the expected time (which is the mean) of delivery.
1.5 marks
Answer: Standard deviation =
(b  a ) 2
= 0.288675 hrs. Therefore, probability of being within one
12
standard deviation of the mean = (0.288675+0.288675)/1 = 0.5774
(b)
The maintenance department of a city’s electric power company finds that it is costefficient to replace all street-light bulbs at once, rather than to replace the bulbs
individually as they burn out. Assume that the lifetime of a bulb is normally (Gaussian)
distributed with a mean of 4000 hours and a standard deviation of 200 hours. If the
department wants no more than 2% of the bulbs to burn out before they are replaced, after
how many hours should all of the bulbs be replaced?
1.5 marks
Answer: Z value for 2 percent or probability 0.02 is -2.05375 from standard normal table.
Thus, -2.05375 = (X – 4000)/200, which gives X = 3589.25. Therefore, approximately all of the bulbs
should be replaced approximately after 3,590 hours.
(c)
How can you check, from data obtained through past observations, that the uncertainty
associated with a process has a normal distribution? Describe in less than 50 words at least
two methods.
1 mark
Answer: (i) By normal probability plot. All data points should lie approximately in a straight line in
the normal probability plot; (ii) By box-and-whiskers plot. The median should be at the centre, the
whiskers of equal length, and the interquartile range about 1.33 times of standard deviation; (iii) By
goodness-of-fit test. Dataset is divided into a number of classes. Relative frequency of each class is
computed and the probability of being in each class is determined from the standard normal curve. A
chi-square test is performed to see if the probabilities are significantly different.
(d)
A firm has monitored the duration of long-distance telephone calls placed by its employees,
to help it decide which long-distance package to purchase. The duration of calls was found
to be exponentially distributed with a mean of five minutes. What proportion of calls last
for more than two minutes?
1 mark
Answer: λ = 1/5 per minute and X = 2 minutes
P(more than 2 minutes) = exp[-0.2(2)] = 0.6703
Question 3
(a)
5 Marks
The probability of success in a trial is 0.65. In 400 trials, what is the probability of
succeeding between 250 and 300 times? Use normal approximation to the binomial
distribution with continuity correction.
1 mark
Answer: Mean = n.p = 400(0.65) = 260, standard deviation = [np(1-p)]0.5 = 9.54
Z1 = (249.5 – 260)/9.54 = -1.153, Z2 = (300.5 – 260)/9.54 = 4.245,
Therefore, required probability P(Z1<X<Z2) = 0.999989 – 0.135529 = 0.8645
(b)
The sign on the lift in a building states ‘Maximum capacity 1120kg or 16 persons’. A
statistics practitioner wonders what the probability is that 16 people would weigh more than
1120kg. If the weights of the people who use the lift are normally distributed with a mean
of 68kg and a standard deviation of 8kg, what is the probability that the statistics
practitioner seeks? [Hint: If you are using the variance equation (5.6) on page 186 of the
textbook (5th edition) assume the covariance to be zero because people’s weights are
independent.]
2 marks
Answer: Mean weight of sixteen persons = 68(16) = 1088
Variance of sixteen people = 16(8)2 = 16(64) = 1024, thus standard deviation = 32
Z = (1120 – 1088)/32 = 1. Therefore, P(X>1120) = 1 – 0.841345 = 0.1587
(c)
The assembly line that produces an electronic component of a missile system has
historically resulted in a 2% defective rate. A random sample of 800 components is drawn.
What is the probability that the defective rate is greater than 4%? Suppose that in the
random sample the defective rate is 4%. What does that suggest about the defective rate on
the assembly line?
2 marks
Answer: Defective rate can be assumed to follow a Poisson distribution. Mean = 800(0.02) = 16,
Variance = 16, thus standard deviation = 4.
X = 800(0.04) = 32, Using normal approximation of the Poisson distribution Z1 = (32-16)/4 = 4.
P(greater than 4%) = P(Z>4) = 1 – 0.999968 = 0.000032, which implies that having such a defective
rate is extremely unlikely.
If the defective rate in the random sample is 4 percent then it is very likely that the assembly line
produces more than 2% defective rate now.
Question 4
(a)
5 Marks
One of the few negative side effects of quitting smoking is weight gain. Suppose that the
weight gain in the 12 months following a cessation in smoking is normally distributed with
a standard deviation of 3 kilograms. To estimate the mean weight gain, a random sample of
13 quitters was drawn and their weight gains in kg are listed below. Determine the 90%
confidence interval estimate of the mean 12-month weight gain for all quitters.
2 marks
8, 12, 4, 1, 7, 11, 9, 6, 5, 9, 3, 4, and 7
Answer: Mean, X = 6.615385, Standard deviation, σ = 3 (known)
Confidence Interval =
(b)
X  Z .05
3
= 6.615385 ± 1.645(0.83205) = (5.25, 7.98)
13
A statistics practitioner working for the Australian Cricket Board wants to supply radio and
television commentators with interesting statistics. He observed several hundred games and
counted the number of times a batsman was out off a no ball. He found there were 373 such
events of which 259 were successful. Estimate with 95% confidence the proportion of all
attempted thefts that are successful.
2 marks
Answer: Proportion of unfair dismissals = 259/373 = 0.69437
Confidence Interval =
(c)
p  Z .025
p(1  p)
= 0.69437 ± 1.96(0.023853) = (0.648, 0.741)
n
If you want to be 99% confident of estimating the population mean to within a sampling
error of ±10 and the standard deviation is assumed to be 50, what sample size is required?
1 mark
Answer: Sample size, n =
Z 2 2 2.575 2  50 2
=
= 165.7656 ≈ 166
10 2
e2
Therefore, the required sample size is 166.