Calculation of Mean and Standard Deviation

The assignment content consists of two main parts: Part A and Part B. Part A deals with the calculation of mean and standard deviation. It provides examples of reducing prices by £12 and £50, and how these changes affect the mean and standard deviation. The second part, Part B, compares the breaking strength of ropes provided by three different suppliers (A, B, and C) based on their mean and standard deviation. It concludes that Supplier B is recommended due to its low variability and high mean compared to Supplier A, which has a higher mean but greater variability.

MATHS

Table of Contents
QUESTION 4.................................................................................................................................................3
2.1 Histogram presentation........................................................................................................................3
QUESTION 5.................................................................................................................................................4
A. Complete the cumulative frequency.....................................................................................................4
B. Cumulative frequency graph.................................................................................................................5
C. (i) Median.............................................................................................................................................5
C. (ii) Lower quartile.................................................................................................................................5
C. (iii) Upper Quartile................................................................................................................................6
C. (iv) The Inter Quartile Range................................................................................................................6
QUESTION 6.................................................................................................................................................7
(A). Calculation of median, lower quartile and upper quartile..................................................................7
B. Calculation of median, lower and upper quartile and inter quartile range............................................8
C. Calculation of median and lower & upper quartile...............................................................................9
QUESTION 7...............................................................................................................................................10
A. (i). Calculation of mean......................................................................................................................10
A. (ii). Calculation of standard deviation.................................................................................................11
(B). Calculation of standard deviation.....................................................................................................11
C. Calculation of mean and standard deviation.......................................................................................12
(a) (i) Calculation of mean & standard deviation....................................................................................12
(d) (i) Calculation of mean and standard deviation.................................................................................13
QUESTION 8...............................................................................................................................................13
A. Calculation of mean and standard deviation.......................................................................................13
(1) Reduction in price by 12....................................................................................................................13
(ii) Reduction in price by 50%.................................................................................................................14
(iii) Mean and standard deviation............................................................................................................14
B. (i) Comparison of breaking strength of the ropes...............................................................................14
B. (ii) Recommendation...........................................................................................................................15
REFERENCES............................................................................................................................................16

QUESTION 4
2.1 Histogram presentation
Frequency distribution
Mass (Gm) Frequency (F)
10-19 6
20-24 4
25-34 12
35-50 18
51-55 8
48
Bin Frequency
10-19 6
20-24 4
25-34 12
35-50 18
51-55 8
10-19 20-24 25-34 35-50 51-55
0
5
10
15
20
6 4
12
18
8
Histogram
Frequency
Bin
Frequency
Above histogram presents that majority of the objects have a mass under the range of 35-
50 gms because its frequency is founded greatest to 18. However, it is founded lowest to 4 for
the mass weight of 20 – 24 gms.

QUESTION 5
A. Complete the cumulative frequency
Cumulative frequency is the continuing total frequency of the given data series.
Time Taken In
Minutes
Frequenc
y Cumulative frequency
Cumulative frequency
%
10-20 1 1 0
20-30 3 1 +3 = 4 5%
30-40 6 4 + 6 = 10 20%
40-50 7 10 + 7 = 17 50%
50-60 3 17 + 3 = 20 85%
20 20 + 20 = 40 100%

B. Cumulative frequency graph
0 10-20 20-30 30-40 40-50 50-60
0%
25%
50%
75%
100%
Cumulative Frequency chart %
C. (i) Median
Median presents 50% value thus divides the data series into two-equal parts, one is below
median (50%) and other one part is above it (Berenson and et.al., 2012).
m = (N/2) Or (∑F/2)
= (20/2) = 10
M = L1 + (N/2-c)/F*i
= 30 + (10-4)/6*10
= 30 + 10
= 40
C. (ii) Lower quartile
Q1 quantifies the 1/4th means 25% values of the given data series, also called lower
quartile (Jaggia and et.al., 2016).

q1 = (N/4)th item
(20/4)
= 5
Q1 = L1 + (N/4-c)/F*i
= 30 + (5-4)/6*10
= 30 + 1.67
= 31.67
C. (iii) Upper Quartile
Q3 quantifies the 3/4th means 75% values of the given data series, also called upper
quartile (Siegel, 2016).
q3 = 3(N/4)th item
= 3(20/4)th item
= 15th item
Q3 = L1 + [3(N/4) – c]/F*i
= 40 + (15 – 7)/10*10
= 40 + 8
= 48
C. (iv) The Inter Quartile Range
IQR is regarded as the excess of upper quartile below lower quartile, computed below:
IQR = Q3-Q1
Q3 = 48
Q1 = 31.67
IQR = (48-31.67)/
= 16.33

QUESTION 6
(A). Calculation of median, lower quartile and upper quartile
Number of players (X) Frequency (F) CF
15 2 2
16 2 4
18 1 5
19 3 8
20 2 10
22 2 12
24 1 13
13
Median (M) = (N+1/2)
= (13+1/2)
= 7
It presented that 50% means half of the registered players in the cricket league is founded
to 7.
It lies in cumulative frequency (CF) of 8, therefore, median value is derived to 19.
Lower quartile (Q1) = (N+1)/4
(13+1)/4
= 14/4
= 3.5
From the results, it can be said that 25% of the players numbered at 3.5 at a lower quartile
value that indicates 1/4th value of the total data series (Newbold, Carlson and Thorne, 2012).
It lies in cumulative frequency (CF) of 4, therefore, lower quartile (Q1) is derived to 16.
Upper quartile (Q3) = 3[(N+1)/4]
3[(13+1)/4]
= 3(14/4)
= 3(3.5)

= 10.5
It lies in cumulative frequency (CF) of 12, therefore, upper quartile (Q3) is derived to 22.
As per the outcome, it can be observed that 75% of the players numbered at 22 at a upper
quartile value that indicates 3/4th value of the given data series.
B. Calculation of median, lower and upper quartile and inter quartile range
Pay (IN GBP) Frequency (F) CF
0-100 4 4
100-200 29 33
200-300 21 54
300-400 38 92
400-500 18 110
500-600 7 117
117
m = (N/2)
= (117/2)
= 58.5
M = L1 + (N/2-c)/F*i
M = £300 + (117/2 – 54)/38*£100
= £300 + (58.5-54)/38*100
= £300 + £11.84
= £311.84
Lower quartile (q1) = (N/4)
= (117/4)
= 29.25
Q1 = L1 + (N/4-c)/F*i
= £100 + (117/4– 4)/29*£100
= £100 + (29.25-4)/29*100
= £100 + (£25.25/£29)*100

= £100 + £87.07
= £187.07
Upper quartile (q3) = 3(N/4)
= 3(117/4)
= 87.75
Q3 = L1 + [3(N/4)-c]/F*i
= £300 + (87.75– 54)/38*£100
= £300 + 88.82
= £388.82
Inter quartile range (IQR) = (Q3 – Q1)/
= (£388.82 - £187.07)
= £201.75
C. Calculation of median and lower & upper quartile
S. No. Test scores
1 4
2 5
3 8
4 9
5 10
6 11
7 12
8 13
9 15
10 15
11 15
12 16
13 17
14 18
15 20
188
Median (M) = Value of (N+1)/2th item
= (15+1)/2th item

= 16/2th item
= value of 8th item
M = 13
Lower quartile (Q1) = (N+1)/4th item
= (15+1)/4th item
= 16/4th item
= Value of 4th item
Q1 = 9
Upper quartile (Q3) = 3(N+1)/4th item
= 3(15+1)/4th item
= 3(16/4)th item
= Value of 12th item
Q3 = 16
QUESTION 7
A. (i). Calculation of mean
Score (X) Frequency (F) FX
1 19 19
2 14 28
3 13 39
4 21 84
5 17 85
6 16 96
100 351
Mean/Average = ∑Fx/∑F
= 351/100
= 3.51
A. (ii). Calculation of standard deviation
By Indirect method through taking assumed mean (A) = 4

Score (X) Frequency (F) FX Dx = X-A Fdx FDx2
1 19 19 -3 -57 171
2 14 28 -2 -28 56
3 13 39 -1 -13 13
4 21 84 0 0 0
5 17 85 1 17 17
6 16 96 2 32 64
100 351 -3 -49 321
σ = √(∑Fdx2/N) – (∑Fdx)2/(∑F)2
= √(321/100 – (-49)2/1002
= √3.21 – (2401/10000)
= √3.21 – 0.24
= √2.97
= 1.72
(B). Calculation of standard deviation
Number of children (X) Number of members (F) Dx = X-A Fdx FDx2
0 9 -3 -27 81
1 4 -2 -8 16
2 6 -1 -6 6
3 5 0 0 0
4 2 1 2 2
5 0 2 0 0
6 1 3 3 9
27 0 -36 114
σ = √(∑Fdx2/N) – (∑Fdx)2/(∑F)2
= √(114/27 – (-36)2/272
= √4.22 – (1296/729)
= √4.22 – 1.78
= √2.44
= 1.56
C. Calculation of mean and standard deviation
Time in
minutes (CI)
Number of
recruits (F) X=L1+L2/2 Fx Dx = x-A FDX FDX2

10-12 3 11 33 -6 -18 108
12-14 14 13 182 -4 -56 224
14-16 30 15 450 -2 -60 120
16-18 16 17 272 0 0 0
18-20 5 19 95 2 10 20
20-22 2 21 42 4 8 32
70 96 1074 -6 -116 504
Mean/Average = ∑Fx/∑F
= 1074/70
= 15.34 minutes
σ = √(∑Fdx2/N) – (∑Fdx)2/(∑F)2
= √504/70 – (-116)2/(70)2
= √7.2 – 13,456/4900
= √7.2-2.75
= √4.45
= 2.11
(a) (i) Calculation of mean & standard deviation
X P(X=x) Xp X2p
1 0.05 0.05 0.05
4 0.12 0.48 1.92
5 0.48 2.4 12
10 0.35 3.5 35
∑p = 1 ∑xp = 6.43 ∑x2p = 48.97
μ = ∑xp = 6.43
Variance (X) = ∑x2p - μ2
= 48.97 – 6.432
= 48.97 – 41.3449
= 7.6251
Standard deviation (σ) = √Variance
= √7.6251

= 2.76
(d) (i) Calculation of mean and standard deviation
X P(X=x) Xp X2p
1 0.15 0.15 0.15
2 0.2 0.40 0.80
3 0.25 0.75 2.25
4 0.4 1.60 6.40
∑p = 1 ∑xp = 2.9 ∑x2p = 9.6
μ = ∑xp = 2.9
Variance (X) = ∑x2p - μ2
= 9.6 – 2.92
= 9.6 – 8.41
= 1.19
Standard deviation (σ) = √Variance
= √1.19
= 1.09
QUESTION 8
A. Calculation of mean and standard deviation
(1) Reduction in price by 12
Current mean = £63
Current standard deviation = £18
If sales price of pair of shoe is reduced by £12, then mean also will be decreased by the
given constant, as follows:
New Mean = £63 - £12 = £51
Standard deviation is not affected by the increase or decrease in the price; therefore, it
remains unchanged from £18.
(ii) Reduction in price by 50%
New Mean = £63 – (£63*50%)

= £63 – £31.5
= £31.5
Standard deviation = £18
(iii) Mean and standard deviation
In the given case, it is stated that later two days, the unsold inventory of pair of shoes
available in the store are valued at a mean price of £70 and standard deviation of £24. Now,
company has decided to reduce the price by £10 and thereafter label the shoe as half price. It
means that previously set price of £110 would be available for sale at £50, calculated as follows:
(£110-£10)/2 = £50
Mean price = £70-£10 = £60
Standard deviation will not be changed and remains still same to that of previous level of
£24. Therefore, new S.D. and average price after discounting will as follows:
Average price of unsold pair of shoes - £60
Standard deviation of unsold pair of shoes - £24
B. (i) Comparison of breaking strength of the ropes
Suppliers Mean Standard deviation
A 856 Kg 390 kg
B 820 kg 42 kg
C 496 kg 28 kg
From the table, it can be seen that according to the central tendency measurement through
average breaking strength, it is founded greater for the supplier A as it is 856 kg which is above
the breaking strength of both the other suppliers to 820 kg and 496 kg. It indicates that as per the
results of mean, A supplier rope’s breaking strength is highest among all. However, on the other
side, dispersion measurement through standard deviations depicts greater volatility and spreader
for the A to 390 kg (Aczel and Sounderpandian, 2002). It clearly reflects that the breaking
strength of different ropes provided by A supplier is significantly different, some of the ropes
breaking strength may be equal to or above mean, whilst other strength is too low. In comparison
to this, B and C supplier standard deviation is comparatively low to 42 and 28 kg shows less
volatility.

B. (ii) Recommendation
On the basis of the findings of average, although breaking strength of A supplier rope is
greatest among all, still due to high variability and scatter in the series, it is considered better to
suggest a friend to bought ropes from the supplier B to set out a climbing holiday. It is because,
its mean shows little difference from the A supplier as it is 820 kg with the less variability.
However, C supplier is not here suggesting because its mean score is very low to 496 kg.

REFERENCES
Books and Journals
Aczel, A.D. and Sounderpandian, J., 2002. Complete business statistics. McGraw-Hill/Irwin.
Berenson, M. and et.al., 2012. Basic business statistics: Concepts and applications. Pearson
higher education AU.
Jaggia, S. and et.al., 2016. Essentials of business statistics: communicating with numbers.
McGraw-Hill Education.
Newbold, P., Carlson, W. and Thorne, B., 2012. Statistics for business and economics. Pearson.
Siegel, A., 2016. Practical business statistics. Academic Press.