Question

Question: In country A, the household incomes are normally distributed, with the mean 25000 dollars and the

variance 10000^2 dollars.

(a) If the poverty level is 10000 dollars, what percentage of the population does not live in poverty?

(b) A new tax law is expected to benefit "middle income" households, those with incomes between 20000 dollars and 30000 dollars. What percentage of the population will benefit from the new tax law?

(c) What is the probability that the mean size of a random sample of 25 chosen households is less than 15000 dollars?

(d) What is the probability that the mean size of a random sample of 400 chosen households is less than 15000 dollars?

(e) Bonus question: why the solution to (c) and (d) are different? what is the intuition behind the difference?

Answer 1

This is a normal distribution question with

a) P(x > 10000.0)=?
The z-score at x = 10000.0 is,

z = -1.5
This implies that
P(x > 10000.0) = P(z > -1.5) = 1 - 0.06680720126885807

b) P(20000.0 < x < 30000.0)=?

This implies that
P(20000.0 < x < 30000.0) = P(-0.5 < z < 0.5) = P(Z < 0.5) - P(Z < -0.5)
P(20000.0 < x < 30000.0) = 0.6914624612740131 - 0.3085375387259869


c) Sample size (n) = 25
Since we know that

P(x < 15000.0)=?
The z-score at x = 15000.0 is,

z = -5.0
This implies that


d) Sample size (n) = 400
Since we know that

P(x < 15000.0)=?
The z-score at x = 15000.0 is,

z = -20.0
This implies that

PS: you have to refer z score table to find the final probabilities.
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