This document contains a student's solutions to exercises in a motorsports engineering course. In exercise 1, the student calculates statistical measures like mean, median, variance, standard deviation, and range for a data set of lap times at a racing circuit. In exercise 2, the student calculates the average, variance and standard deviation of alert signals from an engine RPM sensor over 50 minutes, and solves the statistics questions using Matlab.

1. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
Module: Chassis Development and Telemetry (MSP5001)
Semester 1.
Author: loannis Paraskevas (PhD, CEng)
Franchise Partner Tutor: James Augustus
Email: I.paraskevas@bolton.ac.uk
Email: james.augustus@barnetsouthgate.ac.uk
Student ID: 20277275
Name: OLUWASEGUN BENJAMIN AKINWUMIJU
FDSc: Motorsport Technology.
Date: 02/12/15.
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2. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
Exercise 1 (Total marks:10)
In Silverstone racing circuit, the five fastest laps (in mph) of a F1 car during a race were:
155, 160, 148, 157 and 149.
A. Calculate analytically the mean, median, variance, standard deviation, and range
of the aforementioned set of speeds.
B. Calculate the statistical measures of question a. using Matlab and compare the
results.
SOLUTION TO QUESTION 1.
(i)Mean/ Average = add all the data set
Number of terms.
= 155+160+148+157+149
5
= 769
5
MEAN = 153.8
(ii) Median = the number right in the middle.
To do this you rearrange the data’s.
148, 149, 155, 157, 160.
The Median is 155. (the number in the middle).
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3. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
(iii) Variance = the average of square of the difference between the number
in the data set and the mean.
Value Value Mean Subtracted value Square of Value
of Mean.
148 148–153.8 - 5.8 33.64
149 149–153.8 - 4.8 23.04
155 155–153.8 1.2 1.44
157 157–153.8 3.2 10.24
160 160–153.8 6.2 38.44
Therefore, the variance will be = 33.64+23.04+1.44+10.24+38.44
5
Variance = 106.8
5
Variance = 21.36
(iv). Standard Deviation = √21.36
Standard deviation = 4.62
(V). Range = the different between the highest and the lowest value.
160 – 148
Range = 12
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4. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
EXERCISE 2. (TOTAL MARKS: 15)
In a racing car, the number of incoming alert signals from the engines
revolution per minute (RPM) sensor when the number of revolution of the
engine are more than 8000 in each of 50 successive minutes was recorded
as follows:
1,3,1,2,0,3,1,2,2,3,3,2,0,0,1,3,1,3,0,3,1,0,1,2,1,1,3,2,2,3,1,3,3,1,2,0,0,2,3,3,
0,2,1,0,1,0,3,1,2,0.
A. On average how many alert signals are there per minute? Provide the
analytic solution. (2.5 marks)
A. Calculate the variance and the standard deviation of the above data
set. Provide the analytical solution. (2.5 marks)
B. Solve the question 1a. and 1b. above using Matlab and compare the
results. (10 marks)
SOLUTION TO QUESTION 2.
2a. To calculate for the average, sum up all the value then divide through
by the total number the data’s.
2a. Average =
1+3+1+2+0+3+1+2+2+3+3+2+0+0+1+3+1+3+0+1+0+1+2+1+1+3+2+2+3+
1+3+3+1+2+0+0+2+3+3+0+2+1+0+1+0+3+1+2+0.
= 78
50
= 1.56
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5. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
2b (i). To calculate the variance.
Value Value Mean Subtracted
Value
Square of value
of mean.
0 1.56 ─1.56 2.43
1 1.56 ─0.56 0.31
2 1.56 0.44 0.19
3 1.56 1.44 2.07
Multiply through according to the numbers of data we have.
We have 11 zeros for 0 data’s.
We have 14 data for 1.
We have 11 data for 2.
We have 14 data for 3.
Therefore, for 0 will be. 2.43 x 11 = 26.73
Then for 1 will be 0.31 x 14 = 4.34
Then for 2 will be 0.19 x 11 = 2.09
Then for 3 will be 2.07 x 14 = 28.98
To Calculate for variance now will be = 26.73 + 4.34 + 2.09 + 28.98
50
= 62.14
50
= 1.24
2b(ii). Standard deviation = √1.24
= 1.11
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6. Oluwasegun Benjamin Akinwumiju MSP 5001 07th
December 2015.
2b (i). To calculate the variance.
Value Value Mean Subtracted
Value
Square of value
of mean.
0 1.56 ─1.56 2.43
1 1.56 ─0.56 0.31
2 1.56 0.44 0.19
3 1.56 1.44 2.07
Multiply through according to the numbers of data we have.
We have 11 zeros for 0 data’s.
We have 14 data for 1.
We have 11 data for 2.
We have 14 data for 3.
Therefore, for 0 will be. 2.43 x 11 = 26.73
Then for 1 will be 0.31 x 14 = 4.34
Then for 2 will be 0.19 x 11 = 2.09
Then for 3 will be 2.07 x 14 = 28.98
To Calculate for variance now will be = 26.73 + 4.34 + 2.09 + 28.98
50
= 62.14
50
= 1.24
2b(ii). Standard deviation = √1.24
= 1.11
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