1.20811 BASIC STATISTICS
ASSIGNMENT No. 2 - SEMESTER 1, 2023
TOTAL MARKS: 40
1. The table shows the results of a survey in which 2,944,100 public and 401,900 private school teachers
were asked about their full-time teaching experience.
[10 marks]
a)
b)
c)
d)
e)
Find the probability that a randomly selected private school teacher has 10 to 20 years of full-time
teaching experience.
Given that a randomly selected teacher has 3 to 9 years of full-time experience, find the probability
that the teacher is at a public school.
Are the events “being a public school teacher” and “having 20 years or more of full-time teaching
experience” independent? Explain.
Find the probability that a randomly selected teacher is either at a public school or has less than 3
years of full-time teaching experience.
Find the probability that a randomly selected teacher has 3 to 9 years of full-time teaching experience
or is at a private school.
Solutions
a) Find the probability that a randomly selected private school teacher has 10 to 20 years of full-time
teaching experience.
(2 marks)
111,600
 0.033
P(private and 10 to 20 years) =
3,346 ,000
b) Given that a randomly selected teacher has 3 to 9 years of full-time experience, find the probability
that the teacher is at a public school.
(2 marks)
995 ,800
 0.87
P(public given 3 to 9 years) = P(public / has 3 to 9 years) 
1,150 ,300
c) Are the events “being a public school teacher” and “having 20 years or more of full-time teaching
experience” independent? Explain.
(2 marks)
If the events ‘being a public school teacher’ and ‘having 20 years or more of full-time teaching
experience’ are independent, then:
P(public ) x P(20 years or more )  p(public ) x P(20 years or more given public )
2,944,100 972,900 2.944,100 864,700
x

x
3,346,000 3,346,000 3,346,000 2,944,100
0.256  0.258
Since the left side is not equal to the right side, the events stated are not independent.
d) Find the probability that a randomly selected teacher is either at a public school or has less than 3
years of full-time teaching experience.
(2 marks)
P(public school or has less than 3 years)  P(public )  P(less than 3 years)  P(public and less than 3)
2,944 ,100
204 ,900
177 ,300



3,346 ,000 3,346 ,000 3,346 ,000
 0.89
Therefore, the probability is 0.089.
e) Find the probability that a randomly selected teacher has 3 to 9 years of full-time teaching experience
or is at a private school.
(2 marks)
P(has 3 to 9 years or private )  P(3 to 9 years)  P(private)  P(3 to 9 years and private )
1,150,300
401,900
154,500


3,346,000 3,346,000 3,346,000
 0.42

That is, the probability is 0.42.
2. A corporation has three methods of training employees. Because of time, space, and location, it sends
20% of its employees to location A, 35% to location B, and 45% to location C. Location A has an 80%
success rate. That is, 80% of the employees who complete the course will pass the licensing exam.
Location B has a 75% success rate, and location C has a 60% success rate.
[10 marks]
a) If a person was sent to location C, find the probability that the person failed the exam.
b) If a person has passed the exam, find the probability that the person went to location A.
c) If a person has failed the exam, find the probability that the person went to location B.
Solutions
Lets S for success and F for failure. That is:
S: Success (Passed the exam)
F: Failure (failed the exam)
(3 marks)
First of all, let us present the information in a probability tree diagram. That is:
S
P(A)P(S/A) = 0.2(0.8) = 0.16
0.2
F
P(A)P(F/A) = 0.2(0.2) = 0.04
0.75
S
P(B)P(S/B) = 0.35(0.75) = 0.26
0.25
F
P(B)P(F/B) = 0.35(0.25) = 0.0875
0.6
S
P(C)P(S/C) = 0.45(0.6) = 0.27
0.4
F
P(C)P(F/C) = 0.45(0.4) = 0.18
0.8
A
0.2
0.35
B
0.45
0.2
C
We are asked to find the following probabilities:
a) If a person was sent to location C, find the probability that the person failed the exam.
(1 mark)
That is, given that the person failed the exam, what is the probability that the person was sent to
location C?
The probability is: 𝑃(𝐹|𝐶) = 0.4
b) If a person has passed the exam, find the probability that the person went to location A.
(3 marks)
That is, what is the probability that the person went to location A given that the person has passed the
exam?
P( A / S ) 
P( A) P( S / A)
P( A) P( S / A)

P( S )
P( A) P( S / A)  P( B) P( S / B)  P(C ) P( S / C )
0.16

0.16  0.26  0.27
 0.23
Thus, the probability that the person went to location A given that the person passed the exam is 0.23.
c) If a person has failed the exam, find the probability that the person went to location B.
(3 marks)
That is, what is the probability that the person went to location B given that the person has failed the
exam?
P( B / F ) 
P( B  F ) P( B) P( F / B)
P( B) P( F / B)


P( F )
P( F )
P( B) P( F / B)  P( A) P( F / A)  P(C ) P( F / C )
0.0875

0.0875  0.04  0.18
 0.28
Thus, the probability that the person went to location A given that the person passed the exam is 0.23.
3. Let a random variable X has the function f(x) as follows:
Values of X
0
1
2
3
4
5
6
f(x)
d
3d
5d
7d
9d
11d 13d
[20 marks]
7
15d
8
17d
a. Describe the conditions under which f(x) will be a probability function of the random variable X.
b. Find the value of d such that f(x) is a probability distribution function.
c. Determine the following:
i. P( X  4)
ii. P( X  5)
iii. P(2  X  7)
d. Determine the expected value and standard deviation of X.
Solutions
Values of X
f(x)
0
d
1
3d
2
5d
3
7d
4
9d
5
11d
6
13d
7
15d
8
17d
a. Describe the conditions under which f(x) will be a probability function of the random variable X.
(2 marks)
If f(x) is a probability distribution function of the random variable X in which X takes the values x1, x2, …., xn,
probability for each x value given by the function f(x) will be between 0 and 1, that is 0  f ( x)  1 , and the
n
total probabilities for the random variable X will be 1, that is
 f ( x)  1 .
i 1
(4 marks)
b. Find the value of d such that f(x) is a probability distribution function.
n
Given that one of the conditions is
 f ( x)  1 ,
i 1
n
 f ( x ) d  3d  5d  7d  9d  11d  13d  13d  15d  17d  1
i
i 1
81d  1
1
 0.01235
81
Substituting the value for d gives the f(xi) values as follows:
d
X
f(x)
0
0.012
1
0.037
2
0.062
3
0.086
4
0.111
5
0.136
6
0.160
7
0.185
8 Total
0.210
1
c. Determine the following:
In order to determine the probabilities, we further need to establish the cumulative probabilities for the X
variable. The expanded table in part (b) containing the cumulative probabilities is:
(1 mark)
X
f(x)
F(x)
i.
0
0.012
0.012
1
0.037
0.049
2
0.062
0.111
3
0.086
0.198
4
0.111
0.309
5
0.136
0.444
6
0.160
0.605
7
0.185
0.790
P ( X  4)
P( X  4)  f ( x  0)  f ( x  1)  f ( x  2)  f ( x  3)  f ( x  4)
8 Total
0.210
1
1.000
(2 marks)
 0.012  0.037  0.086  0.111
 0.309
or alternatively, P( X  4)  F ( x  4)  0.309
ii.
P( X  5)
(2 marks)
P( X  5)  f ( x  5)  f ( x  6)  f ( x  7)  f ( x  8)
 0.160  0.185  0.210
 0.556
or alternatively, P( X  5)  1  F ( x  5)  P( x  5)  1  0.444  0.556
iii. P(2  X  7)
P(2  X  7)  f (2  x  7)  f ( x  3)  f ( x  4)  f ( x  5)  f ( x  6)
 0.086  0.111  0.136  0.160
 0.494
or alternatively,
P(2  X  7)  P( X  7)  P( X  2)  F ( x  6)  F ( x  2)  0.605  0.111  0.494
(2 marks)
(7 marks)
d. Determine the expected value and standard deviation of X.
In order to compute the expectation and the standard deviation of X, we also need to expand the table in part c
to include the terms xf (x) and x 2 f ( x) in order to use the expectation formula and variance formula,
respectively. That is:
X
f(x)
F(x)
xf(x)
x2f(x)
0
0.012
0.012
0.000
0.000
1
0.037
0.049
0.037
0.037
2
0.062
0.111
0.123
0.247
3
0.086
0.198
0.259
0.778
4
0.111
0.309
0.444
1.778
5
0.136
0.444
0.679
3.395
6
0.160
0.605
0.963
5.778
7
0.185
0.790
1.296
9.074
8 Total
0.210
1
1.000
1.679
5.48
13.432 34.52
Then, the expectation is:
x  E( X ) 
xf ( x)  5.48

whereas the standard deviation is:
s 2  V ( X )  E ( X 2 )  E ( X ) , where E ( X 2 ) 
2
x
2
f ( x)  34.52 , thus
s 2  V ( X )  34.52  [5.48] 2  34.52  30.047  4.472 .
Thus, the standard deviation is: s  s  4.472  2.11