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The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800?
Most Upvoted Answer
The average weekly food expenditure of a group of families has a norma...
Solution:
Given,
Mean (μ) = Rs 1800
Standard deviation (σ) = Rs 300
Number of families (n) = 5
To find the probability that at least one family has weekly food expenditure in excess of Rs 1800, we can use the concept of complementary probability.
Step 1:
Find the probability that none of the 5 families have weekly food expenditure in excess of Rs 1800.
Let X be the amount spent by a family on food per week. Then, we need to find P(X ≤ 1800) for each family.
Using the standard normal distribution table, we find that the z-score for X = 1800 is (1800-1800)/300 = 0.
Therefore, P(X ≤ 1800) = P(Z ≤ 0) = 0.5.
Since we assume that the weekly food expenditure of each family is independent of the other families, we can use the multiplication rule to find the probability that none of the 5 families have weekly food expenditure in excess of Rs 1800.
P(All 5 families have X ≤ 1800) = P(X ≤ 1800) ^ 5 = 0.5 ^ 5 = 0.03125.
Step 2:
Find the probability that at least one family has weekly food expenditure in excess of Rs 1800.
Using the complementary probability, we can find this probability as:
P(At least one family has X > 1800) = 1 - P(No family has X > 1800)
= 1 - 0.03125
= 0.96875.
Therefore, the probability that at least one family has weekly food expenditure in excess of Rs 1800 is 0.96875.
Conclusion:
The probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800 is 0.96875.
Given,
Mean (μ) = Rs 1800
Standard deviation (σ) = Rs 300
Number of families (n) = 5
To find the probability that at least one family has weekly food expenditure in excess of Rs 1800, we can use the concept of complementary probability.
Step 1:
Find the probability that none of the 5 families have weekly food expenditure in excess of Rs 1800.
Let X be the amount spent by a family on food per week. Then, we need to find P(X ≤ 1800) for each family.
Using the standard normal distribution table, we find that the z-score for X = 1800 is (1800-1800)/300 = 0.
Therefore, P(X ≤ 1800) = P(Z ≤ 0) = 0.5.
Since we assume that the weekly food expenditure of each family is independent of the other families, we can use the multiplication rule to find the probability that none of the 5 families have weekly food expenditure in excess of Rs 1800.
P(All 5 families have X ≤ 1800) = P(X ≤ 1800) ^ 5 = 0.5 ^ 5 = 0.03125.
Step 2:
Find the probability that at least one family has weekly food expenditure in excess of Rs 1800.
Using the complementary probability, we can find this probability as:
P(At least one family has X > 1800) = 1 - P(No family has X > 1800)
= 1 - 0.03125
= 0.96875.
Therefore, the probability that at least one family has weekly food expenditure in excess of Rs 1800 is 0.96875.
Conclusion:
The probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800 is 0.96875.
Community Answer
The average weekly food expenditure of a group of families has a norma...
Given that the distribution is normal .
So the probability that the expenditure of a family is less than the mean by definition, is 0.5.
ie,
The probability of expenditure = Less than the Mean = 0.5
Thus, Probability of all 5 families is less than the mean = 0.5⁵
Thus, Probability of at least one family has weekly food expenditure = 1 - 0.5⁵ = 1 - 0.03125 = 0.96875
Hence, the required probability = 0.96875
So the probability that the expenditure of a family is less than the mean by definition, is 0.5.
ie,
The probability of expenditure = Less than the Mean = 0.5
Thus, Probability of all 5 families is less than the mean = 0.5⁵
Thus, Probability of at least one family has weekly food expenditure = 1 - 0.5⁵ = 1 - 0.03125 = 0.96875
Hence, the required probability = 0.96875
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The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800?
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The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800?.
The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average weekly food expenditure of a group of families has a normal distribution with mean of rs 1800 and standard deviation rs300. What is the probability that out of 5 families belonging to this group, at least one family has weekly food expenditure in excess of rs1800?.
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