A bookshop has a selection of 60 different mathematics books. The number of chapters contained in these books are summarised in the table:

Number of chapters||| 3–5 ||| 6–8 ||| 9 – 11 ||| 12 – 16 ||| 17 – 21 ||| 22 – 30
Number of books||||||||| 4 ||||||||| 6 |||||||| 12 |||||||||||| 14 |||||||||||| 14 ||||||||||||||| 10
1. Find estimates of mean & standard deviation of number of chapters per book
In fact, the values of the mean and standard deviation of the number of chapters per book are 14.7 and 6.1. Use these values for the remainder:
2. For each of the two statements below, state with reasons whether it is definitely true, definitely false or possibly true.
Statement 1 There is at least one outlier below the mean.
Statement 2 There is at least one outlier above the mean.
The number of pages, p, in a book is modelled by p = 20x +15, where x is the number of chapters in the book.
3. Calculate the mean and standard deviation of the number of pages, as given by this model.

Number of chapters||| 3–5 ||| 6–8 ||| 9 – 11 ||| 12 – 16 ||| 17 – 21 ||| 22 – 30
Number of books||||||||| 4 ||||||||| 6 |||||||| 12 |||||||||||| 14 |||||||||||| 14 ||||||||||||||| 10
1. Find estimates of mean & standard deviation of number of chapters per book
In fact, the values of the mean and standard deviation of the number of chapters per book are 14.7 and 6.1. Use these values for the remainder:
2. For each of the two statements below, state with reasons whether it is definitely true, definitely false or possibly true.
Statement 1 There is at least one outlier below the mean.
Statement 2 There is at least one outlier above the mean.
The number of pages, p, in a book is modelled by p = 20x +15, where x is the number of chapters in the book.
3. Calculate the mean and standard deviation of the number of pages, as given by this model.
-
2)
3-5 4 4
6-8 6 10
9-11 12 22
12-16 14 36
17-21 14 50
22-30 10 60
The last column is the cumulative frequencies.
First quartile Q1 = L + w * (n/4 - F) / f, where:
L = lower limit of the interval containing Q1
w = width of the interval containing Q1
n = total frequency
F = cumulative frequency corresponding to the lower limit
f =number of cases in the interval containing Q1
n/4=60/4 = 15
L=9
w=3
n=60
F=10
f =12
Q1 = 9 + 3( 15 -10) /12 = 10.25
Find Q3, the third quartile
3n/4 = 3(60)/4 = 45
L =17
w=4
n=60
F=36
f=14
Q3 = 17 +4(45-36)/14 =19.57
Compute Lowerfence = Q1-1.5(IQR)
Upperfence = Q3+1.5(IQR)
Any data lying outside the bound (L,U) can be considered an outlier
IQR = Q3-Q1 = 19.57 - 10.25 = 9.32
Q1 - 1.5 (IQR) = 10.25 -1.5 (9.32) = -3.73
Q3 +1.5 (IQR) = 19.57 +1.5 (9.32) = 33.55
There are no values outside the range (-3.73, 33.55) , so there are no outliers above or below the mean.
3-5 4 4
6-8 6 10
9-11 12 22
12-16 14 36
17-21 14 50
22-30 10 60
The last column is the cumulative frequencies.
First quartile Q1 = L + w * (n/4 - F) / f, where:
L = lower limit of the interval containing Q1
w = width of the interval containing Q1
n = total frequency
F = cumulative frequency corresponding to the lower limit
f =number of cases in the interval containing Q1
n/4=60/4 = 15
L=9
w=3
n=60
F=10
f =12
Q1 = 9 + 3( 15 -10) /12 = 10.25
Find Q3, the third quartile
3n/4 = 3(60)/4 = 45
L =17
w=4
n=60
F=36
f=14
Q3 = 17 +4(45-36)/14 =19.57
Compute Lowerfence = Q1-1.5(IQR)
Upperfence = Q3+1.5(IQR)
Any data lying outside the bound (L,U) can be considered an outlier
IQR = Q3-Q1 = 19.57 - 10.25 = 9.32
Q1 - 1.5 (IQR) = 10.25 -1.5 (9.32) = -3.73
Q3 +1.5 (IQR) = 19.57 +1.5 (9.32) = 33.55
There are no values outside the range (-3.73, 33.55) , so there are no outliers above or below the mean.