Types of Probability Assigning Methods

STATISTICS

Table of Contents
Question 1........................................................................................................................................3
a. Types of probability assigning methods..................................................................................3
b.i.................................................................................................................................................3
b.ii................................................................................................................................................3
Question 2........................................................................................................................................3
a....................................................................................................................................................3
b...................................................................................................................................................3
Question 3........................................................................................................................................3
a....................................................................................................................................................3
b...................................................................................................................................................3
c....................................................................................................................................................4
d...................................................................................................................................................4
e....................................................................................................................................................4
Question 4........................................................................................................................................4
a....................................................................................................................................................4
b...................................................................................................................................................4
c....................................................................................................................................................5
d...................................................................................................................................................7
Question 5........................................................................................................................................7
i. Derivation of regression equation.............................................................................................7
ii...............................................................................................................................................8
iii..................................................................................................................................................9
Question 6......................................................................................................................................10
a..................................................................................................................................................10
b.................................................................................................................................................11
c..................................................................................................................................................11

Question 1
a. Types of probability assigning methods
The types of probability assigning methods have been described in the form of approaches:
1. Classical approach: Discusses about equally likely events
2. Relative frequency: Assigning of probabilities on the behalf of experimentation or historical
data.
3. Subjective approach: Assignment of probabilities based on subjective judgment.
b.i.
Percentage of the members felt that both job security and salary increment were important
= 0.65 * 0.60 = 0.39
Hence, 39% members feel both job security and salary increment important.
b.ii.
Out of 100%; 39% feel both job security and salary increment important. Hence, remaining feel
at least on these two issues was important. So, percentage of such members is:
= 100 – 39 = 61%
Question 2
a.
The value 12 reflects meals per month, the value of standard deviation = 12 – 10 = 2. And 95%
of values are within 2 standard deviation of the mean (68-95-99.7 rule). Thus, out of 100; 95
students consume more than 12 instant meals per month.
b.
Consumption of instant meals per student = 275 / 25 = 11. And 11 is the indication of 1 standard
deviation (11 – 10 = 1). Where, 68% of values are within 1 standard deviation (68-95-99.7 rule)
and 100% are more than 1 standard deviation of the mean. So, there is 100% chance that a
random sample of 25 students will consume more than 275 instant meals.

Question 3
a.
H0: There is no significance difference between the mean of pre and post installation of the
safety equipment.
H1: There is significance difference between the mean of lost pre and post installation of the
safety equipment.
b.
The suitable test statistics here will be paired t-test. The reason is, paired t-test measures the
mean of two identical samples to find whether there is any difference between the mean of both
samples. If there is any difference in the mean indicates that there is variation in the result due to
installation of safety equipment.
c.
Sample mean (mean difference) = -1.2 (given in the question)
Formula of Paired t-test:
t =
d
√ S 2
n
Where d bar is the mean difference, s² is the sample variance, n is the sample size and t is a
Student t quartile with n-1 degrees of freedom.
Standard deviation = 5 (given in the question)
Sample variance s2 = σ 2 = 52 = 25
N = 50 (given in the question)
t = 1.2
√25 /50
t = 1.69
Degree of freedom = n – 1 = 50 -1 = 49
Critical value of t (0.90; 49) = 1.299 (From t table)
P value = 0.10.