TMA: Fundamentals of Statistics and Probability, MTH219e

This document presents a comprehensive solution to a Statistics and Probability assignment (MTH219e) from Singapore University of Social Sciences. The assignment covers a wide range of statistical concepts and probability models, including graphical displays (dot plots), stem-and-leaf displays, binomial and Poisson distributions, uniform distribution, exponential distribution, geometric distribution, descriptive statistics, and Bayes' theorem. The solution includes detailed explanations, calculations, and interpretations for each question, such as calculating probabilities, means, variances, and percentiles. The assignment addresses real-world scenarios like catapult games, parking tickets, error analysis, and machine malfunctions, providing practical applications of statistical methods. The student demonstrates proficiency in applying statistical formulas, interpreting results, and using tools like Excel and R for data analysis. This document serves as a valuable resource for students studying statistics and probability, offering insights into problem-solving techniques and a deeper understanding of the subject matter.

Running Head: Statistics and Probability
Statistics and Probability Assignment
Student’s Name
Institution Affiliation

Statistics and Probability
Question One
a. Graphical Display
i. The graphical display is dot plot. The similarity is that on both graphs, most values lie
between 135 and 140. The difference is that on the first graph there are outliers
whereas in the second one there are none.
ii. To maximize the probability of having the Ping-Pong balls land within the band,
parents will prefer catapult B to A, since catapult A has more variability than B as
revealed by the presence of outliers in its distribution.
iii. The catapult B should be placed 138 centimeters from the target line. This is because
the 138 is approximate to the target and it’s within the range of most values (135 to
140).
b. Stem-and-Leaf Display
To do Stem-and-Leaf display the data values were arranged in ascending order from the
minimum to maximum (Black, 2009). In our, case from 125 to 164.The display is given below.
Stem Leaf
12 5 6 6 6 6 6 7 9 9 9
13 0 0 0 1 1 2 2 2 2 2 3 5 6 6 7 9
14 0 2 5 5 5 6 6 6 7 8
15 2 5 6
16 4
From the display, it’s clear that most values of the data are on the second stem (13). Also, the
distribution is not asymmetrical as most values lie on the lower side (smaller stems, 12 and 13)

Statistics and Probability
Question Two
a. Parking tickets
The probability that a driver gets at least one parking ticket yearly ( p ) is 0.06
i. Binomial model and Poisson to Binomial distribution
According to Francis and Mousley (2014), Binomial model is given by;
P ( X=k ) = ( n
k ) pk ( 1−p ) n−k
where,
n=sample ¿ 80
p= probability of sucess=0.06
k =number of successful events
Therefore, the probability that 4 drivers get at least one parking ticket will be,
P ( X=4 )=(80
4 ) ( 0.06 )4 ( 0.94 )76
¿ 0.186
Poisson approximation to Binomial
According to Ross (2014), to approximate Poisson to Binomial the terms of binomial and
Poisson model are related by the formula;
p= λ
n
Where,
p=term form binomial model ,
λ=average term ¿ Poissonmodel ,

Statistics and Probability
n=sample ¿ ¿
Therefore,
p= λ
n
0.06= λ
80
λ=4.8
According to Borradaile (2013), from this figure, the Poisson is approximated to Binomial by the
e−λ λk
k ! =(n
k ) ( λ
n )k
(1− λ
n )n−k
Therefore,
P ( X=4 )= e−4,8( 4.8)4
4 ! =0.1820
This probability is less than the one computed using binomial model.
ii. By Poisson approximation to Binomial distribution the probability that at least 3 driver
will get at least one ticket
P ( X ≥3 )=1− {P ( 0 )+ P ( 1 ) + P ( 2 ) }
P ( 0 ) = e−4,8 ( 4.8 ) 0
0 ! =0.0082
P ( 1 ) = e−4,8 ( 4.8 ) 1
1 ! =0.0395
P ( 2 ) = e−4,8 ( 4.8 )2
2 ! =0.0948
P ( 3 ) = e−4,8 ( 4.8 ) 3
3! =0.1517

Statistics and Probability
P ( 4 )= e−4,8 ( 4.8 )4
4 ! =0.1820
Therefore,
P ( X ≥3 ) =1− { P ( 0 ) + P ( 1 ) + P ( 2 ) }
¿ 1− { 0.0082+ 0.0395+ 0.0948 }=1−0.1425
¿ 0.8575
iii. The probability that between 2 to 4 drivers will be get at least one ticket
P ( 2< X < 4 )=P ( X < 4 )−P ( X <2 )
P ( X <2 )=P ( 0 )+ P ( 1 )=0.0082+0.0395=0.047 7
P ( X < 4 )=P ( 0 )+P ( 1 ) + P ( 2 ) + P ( 3 )=0.0082+0.0395+0.0948+ 0.1517=0.294 2
Therefore,
P ( 2< X < 4 )=0.2942−0.0477=0.2465
b. Error of density
Model that can be used is uniform distribution model. According to Grinstead & Snell (2012),
the probability density function;
f ( x )= { 1
b−a , a≤ X ≤ b
0 , elsewhere }, whereb=upper limit=0.025∧a=lower limit=−0.025
Therefore,
f ( x )= 1
0.025−(−0.025) =20
Hence,

Statistics and Probability
f ( x ) = {20 ,−0.025 ≤ X ≤ 0.025
0 , elsewhere }
Probability that error id between 0.010 and 0.015
P ( 0.010< X < 0.015 )=20 ∫
0.010
0.015
1 dx
¿ 20 ( X |0.015
0.010
¿ 20 ( 0.015−0.010 )
¿ 0.1
ii. Probability that error is between -0.012 and 0.012
20 ∫
0.012
0.01 2
1 dx
¿=20 ( X |0.012
0.012
¿ 20 ( 0.012−(−0.012) ) =0.48
iii. Mean and variance of error, X
Mean ( X )= a+b
2 = (−0.025+ 0.025 )
2
¿ 0
Var ( X ) = ( b−a )
2
12 = ( 0.025− ( −0.025 ) )
2
12 = 0.0025
12
¿ 0.00020 8
c. Number of calls
Mean number of calls, Y

Statistics and Probability
Mean ( Y ) =∑
i=0
4
y p y
¿ 0 ( 0.1 ) +1 ( 0.2 ) +2 ( 0.4 ) +3 ( 0.2 ) +4 ( 0.1 )
¿ 2
Var ( Y ) =E ( Y 2 ) − ( E ( Y ) ) 2
E ( Y 2 ) =02 ( 0.1 )+ 12 ( 0.2 ) +22 ( 0.4 ) +32 ( 0.2 ) +42 ( 0.1 )=5.2
Therefore,
Var ( Y )=5.2−22=1.2
d. Unbiased die
When 2 or 4 is obtained the associated value 1 and 0 otherwise as shown below.
Possible
Outcome
1 2 3 4 5 6
Value(X) 0 1 0 1 0 0
P ( value=1 )= 2
6 = 1
3
P ( value=0 )= 4
6 = 2
3
The mean of X;
¿ 0 ( 2
3 )+ 1 (1
3 )= 1
3

Statistics and Probability
Therefore, the mean of X is ¿ 1
3
Question Three
a. Model of Time between sales of an Automobile salesperson
i. The model is Exponential probability model. The probability that t >5
According to Hayter (2012), the exponential distribution this probability is given by the formula;
P ( t> k )=e− λt
P ( t>5 ) =e−0.5 ( 5 ) =0.082 1
j. Mean and variance of t
Mean ( t ) = 1
λ = 1
0.5
¿ 2
Var ( t ) = 1
λ2 = 1
(0.5)2 =4
k. Quantile, q.95 for t
0.95=0.5∫
x
∞
e−0.5t dt
0.95=−1 [ e−0.5 t ] ∞
x
0.95=−1[e¿¿−∞−e−0.5 x ]¿
0.95=e−0.5 t

Statistics and Probability
ln ( 0.95 )=−0.5 t
t= ln ( 0.95 )
−0.5 =0.1026
Therefore, Quantile, q.95 for t is 0.1026
b. Random variable X is modeled by
f ( x ) =4 x3
Median of X,
P ( m )=0.5=4∫
m
1
x3 dx
0.5=4 [ x4
4 ] 1
m
0.5=1−m4
m4 =0.5
m=4
√0.5=0.8409
Therefore, the median is 0.8409
c. Given that X1 , X2 ,∧X3 are independence
Mean and Variance of 2 X1 + X2 +X3 +10
Mean=2 E ¿
¿ 30
Due to independence the

Statistics and Probability
Variance=22 V ¿
¿ 12
d. Exponential distribution with a mean of 4 minutes
Pdf of an exponential distribution is given by;
f ( t )= λ e−λt
From this, the mean is given by;
mean ( t ) =1
λ
1
λ =4
λ= 1
4 =0 .25
Probability that customer is served in less than 3 minutes
P ( t<3 )=1−e−0.25=0.527 6
Then, probability that a person is served in less than 3 minutes in at least 4 days of the next 6
days. This follows a binomial distribution with p=0.5276. This is given by
P=P ( 4 ) + P ( 5 ) + P ( 6 )
P ( 4 ) = ( 6
4 ) ( 0.5276 ) 4 ( 0.4724 ) 2=0.2594
P ( 5 ) = ( 6
5 ) ( 0.5276 ) 5 ( 0.4724 ) 1=0.1159

Statistics and Probability
P ( 6 ) = ( 6
6 ) ( 0.5276 ) 6 ( 0.4724 ) 0=0.0216
Therefore,
P ( X ≥ 4 ) =0.2594+ 0.1159+ 0.0216
¿ 0.3916
Question Four
a. Excel Descriptive statistics: Mean, Median, Standard deviation, Skewness and Range
Weight
Mean 21.075
Median 21.5
Standard Deviation 2.809550
Sample Variance 7.893571
Kurtosis -0.62178
Skewness -0.36911
Range 8.2
Count 8
b. Eighty( 80th) percentile Statistics.
The 80Th percentile is 22.94
c. Descriptive statistics on R command.

Statistics and Probability
Question Five
a. Lab test
Let X = persontested has the disease ,∧Y =results of the test are positive
The probability in question: P( X∨Y )
Information from the question:
P ( Y |X ) =0.95 , P ( Y | X' ) =0.01 , P ( X ) =0.005∧P ( X' ) =0.995
To obtainP( X∨Y ), the Bayes’ Rule need to be employed (Francis and Mousley, 2014).
P ( X |Y ) = P ( Y | X )∗P ( X )
P ( Y |X ) ∗P ( X ) + P ( Y | X' )∗P ( X ' )
¿ 0.95∗0.005
0.95 ( 0.005 ) +0.01 ( 0.995 )
¿ 0.3231
b. Poisson with an average of 3.8 fatalities per month,
x=number of fatalities per mont h

Statistics and Probability
P ( X=k ) = e− λ λk
k ! , k=0,1 , 2 … …
i. Probability of one fatality
P ( X=1 )=3.81∗e−3.8
1 ! =0.085
ii. Expectation of X, E(x) and standard deviation of x
E ( x )=λ=3.8
Var (x )=λ
Standard deviation ( x ) =❑
√ Var (x )=❑
√ 3.8=1.949
c. Automobile machine malfunctioning
Let D=number of defective machine, the probability(p) of been defective is 0.1
This will assume binomial distribution with p=0.1 and n=5
i. P ( X=3 )
¿ (5
3 ) ( 0.2 )3 ( 0.9 )2=0.0081
Hence, the probability that 3 automobile machine are defective is 0.0081
ii. Probability that at least 3 are defective
P ( x ≥ 3 )=1−P ( x <3 )
P ( x<3 )=P ( 0 ) + P ( 1 )+ P

Statistics and Probability
P ( 0 ) = ( 5
0 ) ( 0.2 ) 0 ( 0.9 ) 5=0.5905
P ( 1 ) =( 5
1 ) ( 0.2 ) 1 ( 0.9 ) 4 =0.3281
P ( 2 ) =( 5
2 ) ( 0.2 ) 2 ( 0.9 ) 3=0.0729
Therefore,
P ( x<3 ) =0.5905+0.3281+0.0729=0.9915
Thus,
P ( ≥3 )=1−0.991 5
¿ 0.008 5
iii. Mean and standard deviation of random variable denoting the number of defectives.
mean=np=5∗0.1
¿ 0.5
Var=np ( 1− p )=5∗0.1∗0.9
standard deviation=❑
√var =❑
√5∗0.1∗0.9
¿ 0.6708
d. Geometric distribution with p=0.2
i. Probability mass function of the geometric distribution is given by
P ( X=k ) = ( 1−p ) k p , k =0 , 1, 2 … ..

Statistics and Probability
Mean and variance of X are given by
mean ( x ) =1−p
p =0.8
0.2 =4
Variance ( x ) = 1− p
p2 = 0.8
( 0.2 ) 2 =20
ii. Probabilities
P ( X=4 )= ( 0.8 ) 4 ( 0.2 )=0.128
P ( X >2 )=1−P ( X <2 )=1− { P ( 0 ) + P (1 ) }
P ( 0 )= ( 0.8 )0 ( 0.2 )=0.2
P ( 1 )= ( 0.8 )1 ( 0.2 )=0.16
Therefore,
P ( X >2 ) =1− { 0.2+0.16 ) =0.64

Statistics and Probability
Reference
Borradaile, G. J. (2013). Statistics of earth science data: their distribution in time, space and
orientation. Springer Science & Business Media.
Black, K. (2009). Business statistics: Contemporary decision making. John Wiley & Sons.
Francis, A. and Mousley, B. (2014). Business mathematics and statistics. Andover: Cengage
Learning EMEA.
Grinstead, C. M., & Snell, J. L. (2012). Introduction to probability. American Mathematical
Soc..
Hayter, A. J. (2012). Probability and statistics for engineers and scientists. Nelson Education.
Ross, S. (2014). A first course in probability. Pearson.