The average weekly food expenditure of a group of families has a norma...
Problem:
A group of families has a normal distribution with mean Rs.1800 and standard deviation Rs.300. What is the probability that out of five families belonging to this group at least one family has weekly food expenditure in excess of 2100?
Solution:
Step 1: Find the probability of one family having weekly food expenditure in excess of 2100
We need to find the probability that a single family has a weekly food expenditure in excess of 2100. We can use the standard normal distribution to find this probability.
Z = (X - μ) / σ
Where,
- X = 2100 (weekly food expenditure)
- μ = 1800 (mean weekly food expenditure)
- σ = 300 (standard deviation of weekly food expenditure)
Substitute these values in the formula to get:
Z = (2100 - 1800) / 300 = 0.1
Using the standard normal distribution table, the probability of a family having weekly food expenditure in excess of 2100 is:
P(Z > 0.1) = 0.4602
Step 2: Find the probability of at least one family having weekly food expenditure in excess of 2100
We want to find the probability that out of five families belonging to this group, at least one family has weekly food expenditure in excess of 2100. We can use the binomial distribution to find this probability.
P(X ≥ 1) = 1 - P(X = 0)
Where,
- X = number of families with weekly food expenditure in excess of 2100
- n = 5 (total number of families)
- p = 0.4602 (probability of a family having weekly food expenditure in excess of 2100)
Substitute these values in the formula to get:
P(X ≥ 1) = 1 - P(X = 0) = 1 - C(5,0) * (0.4602)^0 * (1 - 0.4602)^(5-0) = 0.8095
Step 3: Interpretation
The probability that out of five families belonging to this group at least one family has weekly food expenditure in excess of 2100 is 0.8095. This means that there is an 80.95% chance that at least one family will have weekly food expenditure in excess of 2100.