Mathematics Assignment
Added on 2022-11-15
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Statistics and Probability
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Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
16th July 2019
Student Name:
Instructor Name:
Course Number:
16th July 2019
Question 1
Q1 (a)
In the 100m final of the Olympics there are 8 athletes taking part: In how many ways can they be
placed first, second and third?
First 8C1=8 ways
Second 8C2=28 ways
Third 8C3=56 ways
At the end of the race how many ways can the athletes line up for a photograph? 8C3= 56 ways
Q1 (b)
A person randomly selects 5 balls from a bag containing 3 Red, 5 Black and 6 Blue balls. Find
the probability that 2 Black and 3 Blue balls are chosen.
𝑷(𝟐𝒃𝒍𝒂𝒄𝒌 𝒂𝒏𝒅 𝟑 𝒃𝒍𝒖𝒆) = 5
14 × 5
14 × 6
14 × 6
14 × 6
14 = 5400
537824
Q1 (c)
Expand the first four terms of (𝟏 − 𝒙
𝟐)−𝟓
𝟐
(1 − 𝑥
2)−5
2= 1
(1−𝑥
2)
5
2
Expanding (1 − 𝑥
2)2.5= (1)0(-𝑥
2)2.5+2.5(1)1(-𝑥
2)1.5+1.875(1)2(-𝑥
2)0.5+0.3125(1)3(-𝑥
2)-0.5
(1)0(-𝑥
2)2.5+2.5(1)1(-𝑥
2)1.5+1.875(1)2(-𝑥
2)0.5+0.3125(1)3(-𝑥
2)-0.5
= -0.1768𝑥5
2 + 0.8839𝑥3
2 − 1.326𝑥1
2 −0.4419𝑥−1
2
(𝟏 − 𝒙
𝟐)−𝟓
𝟐 = −𝟎. 𝟏𝟕𝟔𝟖𝒙−𝟓
𝟐 + 𝟎. 𝟖𝟖𝟑𝟗𝒙−𝟑
𝟐 − 𝟏. 𝟑𝟐𝟔𝒙−𝟏
𝟐 − 𝟎. 𝟒𝟒𝟏𝟗𝒙𝟏
𝟐
Q1 (d)
For the following eight cars aged 2, 3, 4, 5, 6, 6, 7, and 7 find a 99% Confidence Interval.
Assume that the ages of the cars are normally distributed.
Mean x̄ =2+3+4+ 5+6+6+7 + 7
8 = 40
8 = 5
μ = x̄ ± t sx̄
s=√𝒔𝒖𝒎 (𝒙−𝐱̄)2
𝒏−𝟏 =√(𝟐−𝟓)𝟐+(𝟑−𝟓)𝟐+(𝟒−𝟓)𝟐+(𝟓−𝟓)𝟐+(𝟔−𝟓)𝟐+(𝟔−𝟓)𝟐+(𝟕−𝟓)𝟐+(𝟕−𝟓)𝟐
𝟖−𝟏
=1.852
sx̄ = 𝒔
√𝒏 = 𝟏.𝟖𝟓𝟐
√𝟖 =0.6548
Q1 (a)
In the 100m final of the Olympics there are 8 athletes taking part: In how many ways can they be
placed first, second and third?
First 8C1=8 ways
Second 8C2=28 ways
Third 8C3=56 ways
At the end of the race how many ways can the athletes line up for a photograph? 8C3= 56 ways
Q1 (b)
A person randomly selects 5 balls from a bag containing 3 Red, 5 Black and 6 Blue balls. Find
the probability that 2 Black and 3 Blue balls are chosen.
𝑷(𝟐𝒃𝒍𝒂𝒄𝒌 𝒂𝒏𝒅 𝟑 𝒃𝒍𝒖𝒆) = 5
14 × 5
14 × 6
14 × 6
14 × 6
14 = 5400
537824
Q1 (c)
Expand the first four terms of (𝟏 − 𝒙
𝟐)−𝟓
𝟐
(1 − 𝑥
2)−5
2= 1
(1−𝑥
2)
5
2
Expanding (1 − 𝑥
2)2.5= (1)0(-𝑥
2)2.5+2.5(1)1(-𝑥
2)1.5+1.875(1)2(-𝑥
2)0.5+0.3125(1)3(-𝑥
2)-0.5
(1)0(-𝑥
2)2.5+2.5(1)1(-𝑥
2)1.5+1.875(1)2(-𝑥
2)0.5+0.3125(1)3(-𝑥
2)-0.5
= -0.1768𝑥5
2 + 0.8839𝑥3
2 − 1.326𝑥1
2 −0.4419𝑥−1
2
(𝟏 − 𝒙
𝟐)−𝟓
𝟐 = −𝟎. 𝟏𝟕𝟔𝟖𝒙−𝟓
𝟐 + 𝟎. 𝟖𝟖𝟑𝟗𝒙−𝟑
𝟐 − 𝟏. 𝟑𝟐𝟔𝒙−𝟏
𝟐 − 𝟎. 𝟒𝟒𝟏𝟗𝒙𝟏
𝟐
Q1 (d)
For the following eight cars aged 2, 3, 4, 5, 6, 6, 7, and 7 find a 99% Confidence Interval.
Assume that the ages of the cars are normally distributed.
Mean x̄ =2+3+4+ 5+6+6+7 + 7
8 = 40
8 = 5
μ = x̄ ± t sx̄
s=√𝒔𝒖𝒎 (𝒙−𝐱̄)2
𝒏−𝟏 =√(𝟐−𝟓)𝟐+(𝟑−𝟓)𝟐+(𝟒−𝟓)𝟐+(𝟓−𝟓)𝟐+(𝟔−𝟓)𝟐+(𝟔−𝟓)𝟐+(𝟕−𝟓)𝟐+(𝟕−𝟓)𝟐
𝟖−𝟏
=1.852
sx̄ = 𝒔
√𝒏 = 𝟏.𝟖𝟓𝟐
√𝟖 =0.6548
p(1-0.01
2 ) = 0.995 ,v-1=8-1=7
When p=0.995 and v=7, then t=3.499
μ = x̄ ± tsx̄ = 5 ± 3.499(0.6548)
𝟐. 𝟕𝟎𝟖 ≤ μ ≤ 𝟕. 𝟐𝟗𝟏
Q1 (e)
Explain what Type 1 and Type 2 errors are in Hypothesis Testing.
Type 1 errors-error that occurs when the null hypothesis is rejected instead of being accepted.
Type 2 errors-errors that occur when the null hypothesis is accepted instead of being rejected.
Q1 (f)
Evaluate the determinant of the following matrix:
⌈
6 3 2
4 1 −9
−5 6 −1
⌉
6 3 2 6 3
4 1 -9 4 1
-5 6 -1 -5 6
Determinant=[((6 × 1 × −1) + (3 × −9 × −5) + (2 × 4 × 1))] − [((−5 × 1 × 2) +
(6 × −9 × 6) + (−1 × 4 × 3))]
= (-6+135+8)-(-10-324-12) =137+346=483
Determinant=483
Q1 (g)
Explain the difference between the regression and correlation coefficients.
Regression coefficient is used to show the relationship between the independent and dependent
variable numerically while correlation coefficient is used to show the association strength that
exist between two Variables.
Q1 (h)
Calculate the limit of the following:
lim
𝑥→2
𝑥2 + 3𝑥 − 10
𝑥2 − 8𝑥 + 12
Let’s start by simplifying the expression through factorisation.
x2+3x-10= X2+5x-2x-10
x2 +5x-2x-10=x(x+5)-2(x+5) = (x-2)(x+5)
x2-8x+12= x2-6x-2x+12=x(x-6)-2(x-6)=(x-2)(x-6)
𝐱𝟐+𝟑𝐱−𝟏𝟎
𝐱𝟐−𝟖𝐱+𝟏𝟐 = (𝐱−𝟐)(𝐱+𝟓)
(𝐱−𝟐)(𝐱−𝟔) =(𝒙+𝟓)
(𝒙−𝟔)
2 ) = 0.995 ,v-1=8-1=7
When p=0.995 and v=7, then t=3.499
μ = x̄ ± tsx̄ = 5 ± 3.499(0.6548)
𝟐. 𝟕𝟎𝟖 ≤ μ ≤ 𝟕. 𝟐𝟗𝟏
Q1 (e)
Explain what Type 1 and Type 2 errors are in Hypothesis Testing.
Type 1 errors-error that occurs when the null hypothesis is rejected instead of being accepted.
Type 2 errors-errors that occur when the null hypothesis is accepted instead of being rejected.
Q1 (f)
Evaluate the determinant of the following matrix:
⌈
6 3 2
4 1 −9
−5 6 −1
⌉
6 3 2 6 3
4 1 -9 4 1
-5 6 -1 -5 6
Determinant=[((6 × 1 × −1) + (3 × −9 × −5) + (2 × 4 × 1))] − [((−5 × 1 × 2) +
(6 × −9 × 6) + (−1 × 4 × 3))]
= (-6+135+8)-(-10-324-12) =137+346=483
Determinant=483
Q1 (g)
Explain the difference between the regression and correlation coefficients.
Regression coefficient is used to show the relationship between the independent and dependent
variable numerically while correlation coefficient is used to show the association strength that
exist between two Variables.
Q1 (h)
Calculate the limit of the following:
lim
𝑥→2
𝑥2 + 3𝑥 − 10
𝑥2 − 8𝑥 + 12
Let’s start by simplifying the expression through factorisation.
x2+3x-10= X2+5x-2x-10
x2 +5x-2x-10=x(x+5)-2(x+5) = (x-2)(x+5)
x2-8x+12= x2-6x-2x+12=x(x-6)-2(x-6)=(x-2)(x-6)
𝐱𝟐+𝟑𝐱−𝟏𝟎
𝐱𝟐−𝟖𝐱+𝟏𝟐 = (𝐱−𝟐)(𝐱+𝟓)
(𝐱−𝟐)(𝐱−𝟔) =(𝒙+𝟓)
(𝒙−𝟔)
lim
𝑥→2
𝑥2 + 3𝑥 − 10
𝑥2 − 8𝑥 + 12
= lim
𝑥→2
(𝒙 + 𝟓)
(𝒙 − 𝟔)
lim
𝑥→2
(𝒙+𝟓)
(𝒙−𝟔) =2+5
2−6 = 7
−4 = −1.75
Q1 (i)
Calculate the derivative of following function:
𝑓(𝑥) = ( 1
𝑥3 − 2
𝑥) (𝑥2 + 𝑥)= 𝑥2
𝑥3 + 𝑥
𝑥3 − 2𝑥2
𝑥 − 2
𝑓(𝑥) = 𝑥2
𝑥3 + 𝑥
𝑥3 − 2𝑥2
𝑥 − 2= 1
𝑥 + 1
𝑥2 − 2𝑥 − 2
𝑓(𝑥) 1
𝑥 + 1
𝑥2 − 2𝑥 − 2.
𝒅𝒚
𝒅𝒙 = 𝒇𝟏(𝒙) = − 𝟏
𝒙𝟐 − 𝟏
𝒙𝟑 − 𝟐.
Q1 (j)
Calculate the integral of and simplify the following function:
∫(𝑥3 − 2𝑥2) (1
𝑥 − 5) 𝑑𝑥.
=∫(𝑥3 − 2𝑥2) (1
𝑥 − 5) 𝑑𝑥 = ∫(𝑥2 − 5𝑥3 − 2𝑥 + 10𝑥2) 𝑑𝑥.
= 𝑥3
3 − 5𝑥4
4 − 𝑥2 + 10𝑥3
3 =11𝑥3
3 − 5𝑥4
4 − 𝑥2
=𝟏𝟏𝒙𝟑
𝟑 − 𝟓𝒙𝟒
𝟒 − 𝒙𝟐
QUESTION 2
Q 2(a)
The following table shows the height of 50 male students, measured to the nearest centimeter.
172 180 179 145 151 148 170 152 160 171
156 183 188 159 177 162 153 176 181 190
166 157 149 191 189 150 161 187 179 155
147 171 185 148 189 192 188 173 168 165
178 142 193 163 152 195 197 178 192 198
a) Construct a Histogram for the data using suitable intervals.
𝑥→2
𝑥2 + 3𝑥 − 10
𝑥2 − 8𝑥 + 12
= lim
𝑥→2
(𝒙 + 𝟓)
(𝒙 − 𝟔)
lim
𝑥→2
(𝒙+𝟓)
(𝒙−𝟔) =2+5
2−6 = 7
−4 = −1.75
Q1 (i)
Calculate the derivative of following function:
𝑓(𝑥) = ( 1
𝑥3 − 2
𝑥) (𝑥2 + 𝑥)= 𝑥2
𝑥3 + 𝑥
𝑥3 − 2𝑥2
𝑥 − 2
𝑓(𝑥) = 𝑥2
𝑥3 + 𝑥
𝑥3 − 2𝑥2
𝑥 − 2= 1
𝑥 + 1
𝑥2 − 2𝑥 − 2
𝑓(𝑥) 1
𝑥 + 1
𝑥2 − 2𝑥 − 2.
𝒅𝒚
𝒅𝒙 = 𝒇𝟏(𝒙) = − 𝟏
𝒙𝟐 − 𝟏
𝒙𝟑 − 𝟐.
Q1 (j)
Calculate the integral of and simplify the following function:
∫(𝑥3 − 2𝑥2) (1
𝑥 − 5) 𝑑𝑥.
=∫(𝑥3 − 2𝑥2) (1
𝑥 − 5) 𝑑𝑥 = ∫(𝑥2 − 5𝑥3 − 2𝑥 + 10𝑥2) 𝑑𝑥.
= 𝑥3
3 − 5𝑥4
4 − 𝑥2 + 10𝑥3
3 =11𝑥3
3 − 5𝑥4
4 − 𝑥2
=𝟏𝟏𝒙𝟑
𝟑 − 𝟓𝒙𝟒
𝟒 − 𝒙𝟐
QUESTION 2
Q 2(a)
The following table shows the height of 50 male students, measured to the nearest centimeter.
172 180 179 145 151 148 170 152 160 171
156 183 188 159 177 162 153 176 181 190
166 157 149 191 189 150 161 187 179 155
147 171 185 148 189 192 188 173 168 165
178 142 193 163 152 195 197 178 192 198
a) Construct a Histogram for the data using suitable intervals.
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