Statistics and Probability Exam: Data Analysis and Probability

This document contains solutions to a Statistics and Probability online exam. The solutions include calculations and explanations for various statistical concepts. Question 1 covers data analysis, including calculating the mean, median, standard deviation, quartile deviation, constructing a box plot, and creating a frequency distribution table. Question 2 addresses binomial distribution problems. Question 3 explores probability calculations. Question 4 delves into marginal distributions, covariance, and expected values. Question 5 involves a probability calculation. Question 6 focuses on sales data and probability. Question 7 discusses the effect of a course on broker returns. The document provides a comprehensive overview of statistical concepts and problem-solving techniques.

Statistics and
Probability[online exam]

TABLE OF CONTENTS
QUESTION 1..................................................................................................................................3
QUESTION 2..................................................................................................................................7
QUESTION 3..................................................................................................................................7
QUESTION 4..................................................................................................................................7
QUESTION 5..................................................................................................................................7
QUESTION 6..................................................................................................................................7
QUESTION 7..................................................................................................................................8

QUESTION 1
Data given is
55 54 47 55 58
56 61 48 51 44
45 59 58 64 57
48 55 45 66 74
Arranging the data in the ascending order we get,
44, 45, 45, 47, 48, 48, 51, 54, 55, 55, 55, 56, 57, 58, 58, 59, 61, 64, 66, 74,
a. Mean of the data values, μ = Sum of all the values / number of values
= 1040 / 20
= 52
x x-μ ∣x−μ∣2
44 -8 64
45 -7 49
45 -7 49
47 -5 25
48 -4 16
48 -4 16
51 -1 1
54 2 4
55 3 9
55 3 9
55 3 9
56 4 16
57 5 25
58 6 36
58 6 36
59 7 49
61 9 81
64 12 144
66 14 196
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74 22 484
Total (∑) 1318
Number of data values = 20
Q2 = Median of the data set
Since n is even, median can be calculated as
Q2 = (1/2) [(n/2)th observation and (n/2 + 1)th observation
= (1/2) [10th observation and 11th observation]
= (1/2) [55+55]
= 110/2
= 55
Therefore, the median is 55.
b. Standard deviation: In order to calculate the standard deviation, mean of the data is
calculated which was obtained as 52.
SD = √(∑∣x−μ∣2 / N)
= √ (1318 / 20)
= √ 65.9
= 8.12
Quartile Deviation (QD) : In order to calculate QD, following steps need to be followed:
Lower half of the data is 44, 45, 45, 47, 48, 48, 51, 54, 55, 55
The median of this is to be found in the same manner
Q1 = (1/2) [(n/2)th observation and (n/2 + 1)th observation
= (1/2) [5th observation and 6th observation]
= (1/2) [48+48]
= 96/2
= 48
The median of the upper half is also calculated in the similar manner as follows:
55, 56, 57, 58, 58, 59, 61, 64, 66, 74,
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Q3 = (1/2) [(n/2)th observation and (n/2 + 1)th observation
= (1/2) [5th observation and 6th observation]
= (1/2) [58+59]
= 117/2
= 58.5
Quartile deviation = (Q3 – Q1) / 2
= (58.5 – 48) / 2
= 10.5 / 2
= 5.25
c. Box plot
The box plot can be constructed by making use of five values which include the
minimum value, the median, the first quartile, the third quartile and the maximum value.
Minimum value 44
First quartile 48
Median 55
Third quartile 58.5
Maximum value 74
d. The range is to be found as
Highest – lowest value = 74 – 44 = 30
Dividing this by the reasonable class which is 5.
30 / 5 = 6
Rounding this gives us 6.
Frequency distribution table
x Tally marks Frequency (f)
44 | 1
45 || 2
47 | 1
48 || 2
51 | 1
5

54 | 1
55 ||| 3
56 | 1
57 | 1
58 || 2
59 | 1
61 | 1
64 | 1
66 | 1
74 | 1
e.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0
10
20
30
40
50
60
70
80
x
Frequency (f)
f. Percentage of those checked drove faster than average i.e., 52 is obtained as 13/ 20 = 0.65
= 65%.
QUESTION 2
a. The variable X is binomially distributed when randomly selected course participants
will complete the course.
b. 45 out of 79 employees are likely to complete the course.
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c. 25% probability is there that 20 participants will complete the course.
d. 70% is the probability that more than 30 employees will complete the course.
QUESTION 3
a. Probability = 342/ 1207 = 28% probability
b. Probability = 24/1207 = 1.9% probability
c. Probability = 24/543 = 4.4% probability
QUESTION 4
a. The marginal distribution of X is 0.2 and of y is 0.1.
b. The covariance can be found between the two variables x and y.
Cov(X,Y) = Σ E((X-μ)E(Y-ν)) / n-1
Where x is a random variable and n is the number of values.
c. W = X + Y = 1 after adding all the values of x and y.
d. The expected value of W is 1 and the variance is found to be 0.2.
QUESTION 5
Probability = 3/9 = 0.33 which means 33%.
3 is the number of people randomly chosen and 9 are the people who have not taken drugs.
QUESTION 6
X sales per day and the mean is given as 27.3. Standard deviation is given as 4.2.
50 – (27.3)(4.2) = -10,924.6
Interval can be 0-11000.
The probability will be 25%.
QUESTION 7
Course will have a positive effect on the total return of the broker because the before
return are lesser than the after return in cases of majority og the brokers which will lead to
profitability.
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