Questions:
(b) Use the relative frequency approach to find the probability distribution for the Occupational category.
(c) Draw the bar chart for the probability distribution of Occupational category.
(d) Define the probability distribution based on part (b).
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(x) | 0.14 | 0.26 | 0.3 | 0.15 | 0.08 | 0.07 |
i. Find the probability of exactly two
ii. Find the probability more than two
iii. Find the probability at least three.
Answers:
OCCUPATION | Count |
6 | 60 |
Thus, we observe that in the given sample, all the people belong to the occupational category “Others”.
b. Relative frequency is a good approach to compute the probability of an event. Suppose the frequency of an event is f and the total frequency is n. Then the relative frequency of the event is (frequency of the event)/(Total frequency) =f/n.
In this case, we have only one occupational category and its relative frequency
= frequency/60
=60/60
=1
Thus the probability distribution of occupational category is shown below:
OCCUPATION | Count | Probability |
6 | 60 | 1 |
c. The bar chart for the probability distribution of Occupational category is given below:
In the above diagram, 1 actually denotes Occupational Category “6” and all the samples units belong to the category “Others”.
d. Considering the entire population, let X denote the Occupational Category of a person. Then X can take the values 1, 2, 3, 4, 5 and 6. According to our data given in task 1, we have a random sample of size 50 from the population and the corresponding probabilities of X are calculated according to the relative frequency method.
For example P(X=1) = (number of people with Occupational Category 1)/50
In this way we compute all the probabilities for other values of X and the probability distribution table is shown below:
X | 1 | 2 | 3 | 4 | 5 | 6 |
P(x) | 0.1 | 0.18 | 0.26 | 0.24 | 0.02 | 0.2 |
i. Based on the probability distribution given in the above probability distribution table, we compute some of the probabilities
ii. Probability of exactly two
=P(X=2)
=0.18 (given in the above table)
iii. Probability of more than two
=P(X>2)
=P(X=3) + P(X=4) + P(X=5) +P(X=6)
=0.26+0.24+0.02+0.2
=0.72
Probability of atleast three
=P(X>=3)
=P(X=3) + P(X=4) + P(X=5) +P(X=6)
=0.26+0.24+0.02+0.2
=0.72