Statistics Problems - Module Name, Example University, Semester 1

This document presents solutions to various statistics problems. The solutions cover topics such as calculating expected return on investment, determining the probability of passing a test based on the number of correct answers, calculating the probability of mean expenditure falling within a specific range, and determining the IQ score representing the top 25% of a population. The assignment provides detailed explanations and calculations for each problem, offering a comprehensive understanding of the concepts. This assignment is a valuable resource for students seeking to improve their understanding of statistical principles and problem-solving skills.

Running Head: STATISTICS PROBLEMS
Statistics Problems
Name of the Student
Name of the University
Author Note

1STATISTICS PROBLEMS
Answer 1
On investing 20,000 dollars, the possible returns and their respective probabilities are
given by the following table:
Returns (xi) Probabilities (pi) Xi*pi
0 0.4 0
50000 0.4 20000
70000 0.15 10500
100000 0.05 5000
The expected return on the investment can be given by the following formula:
E ( X ) =∑
1=1
n
xi pi
¿ ∑
i=1
4
xi pi
¿ x1 p1 + x2 p2 +x3 p3 +x4 p4
¿ ( 0∗0.4 )+ ( 50,000∗0.4 ) + ( 70,000∗0.15 )+ ( 100,000∗0.05 )
¿ 0+20,000+10,500+ 5000=35500.
Thus, the required expected return of the investment is $35,500.

2STATISTICS PROBLEMS
Answer 2
Total number of questions in a test = 5
Let A be the event “number of correct answers” and B be the event “number of possible
answers”.
Minimum number of correct answers to pass the test = 4
Therefore, the probability of getting 4 answers correct = P ( A )= 4
5
The number of possible answers for each question = 3
Therefore, the probability of getting each question correct = P ( B )= 1
3
Thus, the probability of the student passing the test = P ( A )∗P ( B ) = ( 4
5∗1
3 ) = 0.27

3STATISTICS PROBLEMS
Answer 3
The mean expenditure incurred by a student is $8000
The standard deviation incurred by a student = $800
Sample size = 64
For the mean expenditure of the student to be between $7800 and $8120, the following
probability has to be calculated:
P ( 7800≤ X ≤ 8120 )=P ( X ≤ 8120 )−P ( X ≤7800 )
¿ P
( X −8000
800
√64
≤ 8120−8000
800
√64 )−P
( X−8000
800
√64
≤ 7800−8000
800
√64 )
¿ P
( X −8000
800
8
≤ 120
800
8 )−P
( X−8000
800
8
≤ −200
800
8 )
¿ P ( X −800
100 ≤ 120
100 )−P ( X −8000
100 ≤ −200
100 )
¿ P ( Z ≤1.2 ) −P ( Z ≤−2 )
¿ P ( Z ≤1.2 ) + P ( Z ≤2 )−1
¿ 0.8849+0.9772−1
¿ 0.8621

4STATISTICS PROBLEMS
Answer 4
IQ is normally distributed.
Mean IQ (μ) = 100
Standard deviation of the IQ (σ) = 15.
The top 25 percent of the score will be those scores above 100−25
100 =0.75.
The z-score corresponding to 0.75 = 0.7733. (Tabulated value of z)
The corresponding score which will represent the top 25 percent of the population is given by
X =z∗σ +μ
¿ 0.7733∗15+100
¿ 111.6
An IQ more than 111.6 will represent the top 25 percent of the population.