Stats Pack: Corrections Page 1

Stats Pack
Corrections to 2014 study material

Comment

This document contains details of any errors and ambiguities in the Stats Pack study
materials for 2014 that have been brought to our attention. We will incorporate these
changes in the study material each year. We are always happy to receive feedback from
students, particularly details concerning any errors, contradictions or unclear statements
in the courses. If you have any such comments on this course please email them to
StatsPack@bpp.com.

Important note

This document was last revised on 3rd June 2014. The date on which any corrections
have been added is noted at the start of each section.

The Actuarial Education Company © IFE: 2014 Examinations

Page 2 Stats Pack: Corrections

Chapter 1 (updated on 22nd April 2014)

Page 24

There is an error in the first paragraph after the stem and leaf diagram. It should read
“whereas type B claims are located between $300 and $400”

Replacement pages can be found at the end.

Page 37

The axes for the bar chart in solution 1.6 are the wrong way round. “Frequency” should
be on the vertical axis and “Mock Results” should be on the horizontal axis.

Replacement pages can be found at the end.

Chapter 2 (updated on 30th January 2014)

P2.7 Page 34 and page 47

There is a typo in the table on the question and the solution. The figure of 3300 should
be £300.

Replacement pages can be found at the end.

Chapter 7 (updated on 30th January 2014)

Solution 7.9 Page 58

The second row of the table it should read P(V = v) not F (v) .

Replacement pages can be found at the end.

Chapter 7 (updated on 30th January 2014)

Solution P7.10(ii) Page 80

There is a typo in the first line of part (ii) of the solution. It should read “using
var(aX + b) = a 2 var( X ) ”.

Replacement pages can be found at the end.

© IFE: 2014 Examinations The Actuarial Education Company

Stats Pack: Corrections Page 3

Chapter 8 (updated on 2nd June 2014)

Mode Page 27

There is a typo in the first line under the graph of the Poisson distribution. It should
read “It is easy to see that 1 and 2 are both modes as these have the greatest probability
of occurring.”

Replacement pages can be found at the end.

The Actuarial Education Company © IFE: 2014 Examinations

 All study material produced by ActEd is copyright and is sold
for the exclusive use of the purchaser. The copyright is owned
by Institute and Faculty Education Limited, a subsidiary of
the Institute and Faculty of Actuaries.

Unless prior authority is granted by ActEd, you may not hire

out, lend, give out, sell, store or transmit electronically or
photocopy any part of the study material.

You must take care of your study material to ensure that it is

not used or copied by anybody else.

Legal action will be taken if these terms are infringed. In

addition, we may seek to take disciplinary action through the
profession or through your employer.

These conditions remain in force after you have finished using

the course.

© IFE: 2014 Examinations The Actuarial Education Company

 Stats Pack-01: Statistical diagrams Page 23

Question 1.15

Use your cumulative frequency curve from Question 1.14 to estimate:

(i) how many policyholders are aged 32 or less

(ii) the age under which 75% of the policyholders lie.

3.6 Boxplot

A boxplot (also called a box and whisker plot) is another way of showing data:

25% of data 25% of data 25% of data 25% of data

lowest lower median upper highest

value quartile quartile value
M
Q1 Q3

The rectangle (box) in the middle represents the middle 50% of the data (between the
values that are a ¼ and ¾ of the way through the data). The lines (whiskers) extend
from the box to the smallest and largest values. The diagram also shows the middle
value (called the median).

A boxplot is particularly effective when comparing two sets of data, however to draw
the diagram we need to calculate the median and the quartiles. Since the median will be
covered in Chapter 2 and the quartiles will be covered in Chapter 3 we will deal with
this type of diagram at the end of Chapter 3.

In the exam it is expected that you would draw a boxplot accurately on graph paper.

The Actuarial Education Company © IFE: 2014 Examinations

 Page 24 Stats Pack-01: Statistical diagrams

4 Using diagrams to compare data

Once we have drawn our diagrams we can use them to interpret the patterns in the data
or compare two or more data sets. In Subject CT3 we will be looking at three features
of any data set: the location, the spread and the skewness.

4.1 Location

The location of a data set is simply where the data is located – ie where is the centre of
the data or about what values is it grouped. In everyday language you may use
‘average’ to describe the location.

The stem and leaf diagrams below show the claim amounts (in $’s) under two different
types of insurance:

Type A Type B

0 2 7 0 8
1 1 1 3 6 8 9 1 0 2 3
2 3 4 4 4 7 2 1 4 6 8
3 0 5 3 2 3 3 6 9 9
4 1 4 0 1 5
5 2 5 4

Key: 2|5 represents $250

Type A claims are mostly located between $100 and $200 whereas type B claims are
located between $300 and $400. So we could say the type B claims are greater on
average than type A claims.

In Chapter 2, we will use the mean, median and mode to measure the location of a set of
data.

© IFE: 2014 Examinations The Actuarial Education Company

Stats Pack-01: Statistical diagrams Page 37

Solution 1.5

The cumulative frequency table is:

Cumulative
Time (t)
Frequency
t  0.5 2
t 1 7
t2 14
t 5 26
t  10 30

Or we could use “up to 0.5 mins”, “up to 1 min”, etc as the groups.

Solution 1.6

Putting this data into a frequency table:

Mock result Frequency

68 1
69 2
70 3
71 6
72 5
73 0
74 2
75 1

It is now easy to draw the bar chart:

7
6
5
Frequency

4
3
2
1
0
68 69 70 71 72 73 74 75
Mock Results

The Actuarial Education Company © IFE: 2014 Examinations

Page 38 Stats Pack-01: Statistical diagrams

Solution 1.7

frequency
(i) Using frequency density  we get:
class width

Claim amount (x) Frequency Frequency density

0  x  250 60 60  250  0.24
250  x  500 75 75  250  0.3
500  x  1,000 50 50  500  0.1
1,000  x  2,000 40 40  1,000  0.04
2,000  x  5,000 30 30  3,000  0.01

(ii) The histogram is:

0.3

0.25
Frequency density

0.2

0.15

0.1

0.05

claim amount (£)

0 1,000 2,000 3,000 4,000 5,000

© IFE: 2014 Examinations The Actuarial Education Company

 Stats Pack-02: Sample calculations 1 Page 33

P2.4 Subject 101, September 2000, Q2 (part)

Consider a random sample of 47 white-collar workers and a random sample of 24 blue-

collar workers from the workforce of a large company. The mean salary for the sample
of white-collar workers is £28,470; whereas the mean salary for the sample of blue-
collar workers is £21,420.

Calculate the mean of the salaries in the combined sample of 71 employees. [1]

Section 3: Sample median

P2.5 Subject C1, September 1995, Q1 (adjusted)

A random sample of 15 motor windscreen claim amounts (in £) is given by:

121 107 139 72 123

114 215 156 100 136

169 89 115 153 111

What is the median claim amount? [2]

P2.6 Subject 101, September 2001, Q1 (part)

Data were collected on 100 consecutive days for the number of claims, x, arising from a
group of policies. This resulted in the following frequency distribution:

x 0 1 2 3 4 ≥5
f 14 25 26 18 12 5

Calculate the median for these data. [1]

The Actuarial Education Company © IFE: 2014 Examinations

 Page 34 Stats Pack-02: Sample calculations 1

P2.7 Subject C1, September 1997, Q9 (part)

The table below shows a grouped frequency distribution for 100 claim amounts on a
certain class of insurance policy.

Claim Amount Frequency

under £100 4
£100 – 149.99 10
£150 – 199.99 25
£200 – 249.99 30
£250 – 299.99 15
£300 – 349.99 12
£350 – 399.99 4
£400 or over 0

Determine an approximate value for the median of these claim amounts. [4]

Section 5: Location and skewness

P2.8 Subject C1, Specimen 1993, Q2

For a particular class of insurance policy the distribution of claim amounts is positively
skewed. Which of the following statements about the claim amount distribution is true?

A mode > median > mean

B mean > median > mode

C median > mode > mean

D mean > mode > median [2]

© IFE: 2014 Examinations The Actuarial Education Company

 Stats Pack-02: Sample calculations 1 Page 47

P2.5 First of all we need to put the claim amounts in order:

72 89 100 107 111 114 115 121 123 136 139 153 156 169 215

The median is the ½ ¥ 15 + ½ = 8th value which is £121.

P2.6 The median is the ½ ¥ 100 + ½ = 50½ th value. So counting through the frequencies:

x 0 1 2 3 4 ≥5
f 14 25 26 18 12 5

14 values

39 values

65 values

So the 50½th value is 2 claims.

P2.7 The median is the ½ ¥ 100 = 50 th value. So counting through the frequencies:

Claim Amount Frequency 4

under £100 4 14
£100 – 149.99 10
£150 – 199.99 25
£200 – 249.99 30 39
£250 – 299.99 15
£300 – 349.99 12 69
£350 – 399.99 4
£400 or over 0

The 50th value is the 11th value in the £200 - £249.99 group. Using interpolation, we
get the median to be:

11
200 + ¥ 49.99 = £218.33
30

The Actuarial Education Company © IFE: 2014 Examinations

 Page 48 Stats Pack-02: Sample calculations 1

P2.8 In Section 5 we had the following diagram:

mode
median
mean

Hence, we can see that answer B is correct.

P2.9 Using the sum of the fifty amounts given in the question:

£92, 780
x= = £1,855.60
50

The median is the ½ ¥ 50 + ½ = 25½ th value. Counting through the leaves, we see that
this lies between the 3 and the 4 on the 17 stem (ie between 1,730 and 1,740). Hence,
the median is £1,735.

© IFE: 2014 Examinations The Actuarial Education Company

Stats Pack-07: Discrete random variables Page 57

Solution 7.8

(i) The graph will be:

FW (w)

0.8

0.6

0.4

0.2

0
0 1 2 3 4 5 6 w

(ii) From this we get the cumulative distribution function to be:

Ï0 w<2
Ô
ÔÔ0.2 2£w<4
FW ( w) = Ì
Ô0.7 4£w<5
Ô
ÔÓ1 5£ w

(iii) Either reading off the graph or using the CDF we get:

FW (1) = 0

FW (4.5) = 0.7

FW (10) = 1

The Actuarial Education Company © IFE: 2014 Examinations

Page 58 Stats Pack-07: Discrete random variables

Solution 7.9

We can solve this by obtaining the probability function or by using the cumulative
function directly.

Obtaining the probability function

First we notice that the CDF jumps up at x = 1, 2,3 and 4 . To obtain the probabilities of
obtaining each of these values, we simply subtract the cumulative probabilities:

v 1 2 3 4
P(V = v) 0.216 0.432 0.288 0.064

We can now obtain the probabilities:

(i) P(V = 2) = 0.432

(ii) P (V > 1) = P (V = 2) + P (V = 3) + P (V = 4) = 0.432 + 0.288 + 0.064 = 0.784

(iii) P (V < 3) = P (V = 1) + P(V = 2) = 0.216 + 0.432 = 0.648

Using the cumulative distribution function directly.

(i) Using the fact that subtracting cumulative probabilities gives the original
probabilities:

P (V = 2) = FV (2) - FV (1) = 0.648 - 0.216 = 0.432

(ii) Since probabilities of all possible values sum to 1, we get:

P (V > 1) = 1 - P(V £ 1) = 1 - FV (1) = 1 - 0.216 = 0.784

(iii) Reading directly from the cumulative distribution function:

P (V < 3) = 0.648

© IFE: 2014 Examinations The Actuarial Education Company

 Stats Pack-07: Discrete random variables Page 79

P7.9 The mean of C is given by:

E (C ) = E (7.00 + 0.0742 N )

= 7.00 + 0.0742 E ( N ) using E (aX + b) = aE ( X ) + b

= 7.00 + 0.0742 ¥ 600 since E ( N ) = 600

= 51.52

The variance of C is given by:

var(C ) = var(7.00 + 0.0742 N )

= 0.07422 var( N ) using var(aX + b) = a 2 var( X )

= 0.07422 ¥ 250 since var( N ) = 250

= 1.37641

The Actuarial Education Company © IFE: 2014 Examinations

 Page 80 Stats Pack-07: Discrete random variables

P7.10 (i) E (3 + 6U ) = 3 + 6 E (U ) using E (aX + b) = aE ( X ) + b

= 3+6¥8 since E ( N ) = 600

= 51

(ii) var(8 - 2U ) = ( -2) 2 var(U ) using var(aX + b) = a 2 var( X ) + b

= 4 ¥ 32 since var(U ) = 32
= 36

So the standard deviation is 36 = 6 .

Alternatively, we could have used sd (aX + b) = a ¥ sd ( X ) :

Ê U - 8ˆ
(iii) var Á
Ë 3 ˜¯
= var ( 13 U - 83 )
( 13 )
2
= var(U ) using var(aX + b) = a 2 var( X )

= 19 ¥ 32 since var(U ) = 32

=1

(iv) E (U 2 - 4U + 7) = E (U 2 ) - 4 E (U ) + 7splitting up the expectation

= E (U 2 ) - 4 ¥ 8 + 7 since E (U ) = 8

= E (U 2 ) - 25

To work this out we need E (U 2 ) . We don’t have the distribution, so we can’t

work it out from first principles. We use the trick of rearranging the variance
formula:

var(U ) = E (U 2 ) - E 2 (U )

fi E (U 2 ) = var(U ) + E 2 (U ) = 32 + 82 = 73

Hence:

E (U 2 - 4U + 7) = 73 - 25 = 48

© IFE: 2014 Examinations The Actuarial Education Company

Stats Pack-08: Discrete distributions Page 27

Median

There is no easy way to get the median value other than counting through the
probabilities to find which value is half way through the distribution. For example,
when X ~ Poi (2) we have:

x 0 1 2 3 4 5 …
P(X = x) 0.135 0.271 0.271 0.180 0.090 0.036 …

0.135 0.135  0.271  0.271  0.677

0.135  0.271  0.406 so 0.5 is in here!

So we can see that the median is 2.

Mode

The mode is the value that has the greatest probability. Looking at either the probability
distribution of X ~ Poi (2) given above or its graph below:

0.3

0.2
P(X=x)

0.1

0
0 1 2 3 4 5 6 7 8 9 10
x

It is easy to see that 1 and 2 are both modes as these have the greatest probability of
occurring.

Question 8.15

Calculate the median and mode of X ~ Poi (1) .

The Actuarial Education Company © IFE: 2014 Examinations

 Page 28 Stats Pack-08: Discrete distributions

In summary:

Poisson distribution, Poi (l )

lx
P ( X = x) = e-l x = 0,1, 2, 
x!

E( X ) = l

var( X ) = l

These results are given in the Tables on page 7 and so do not need to be memorised.

4.4 Probabilities of a Poisson distribution

We can use the probability function to calculate probabilities. Suppose X ~ Poi (3) , our
probability function is:

3x -3
P ( X = x) = e x = 0,1, 2, 
x!

To calculate a single probability, P ( X = 2) , we just substitute the value into our

probability function:

32 -3 9 -3
P ( X = 2) = e = e = 0.22404
2! 2

What about calculating P( X ≥ 2) ? Since X ~ Poi (3) can take values 0,1, 2,  this
means:

P ( X ≥ 2) = P( X = 2) + P( X = 3) + P( X = 4) + 

Eek! There’s no way we can work it out this way! So we need to use the fact that
probabilities of all the values X can take sum to 1:

P ( X < 2) + P( X ≥ 2) = 1 fi P( X ≥ 2) = 1 - P( X < 2)

© IFE: 2014 Examinations The Actuarial Education Company