Question

Suppose the household incomes for Kamloops are normally distributed with mean $62,000 and standard deviation $6,800.

a) If 50 households are randomly selected, find the probability that their mean household income is below $64,000.

b) If households with the bottom 20% of incomes qualify for a special tax cut, what is the maximum income required to qualify for the tax cut?

Answer 1

solution:

the given information as follows:

mean income = = $ 62,000

standard deviation = = $6,800

a)

sample size = n = 50

we have to find the probability that mean household income is below 64,000 = P( < 64000)

calculating the z score

P( < 64000) = value of z to the left of 2.08 from the z table = 0.9812

b)

it is given that the bottom 20% household income get a special tax cut.

so we will find the z score with 20% area to the left of the distribution

so z score with 0.20 area to the left = -0.84

so maximum income required to get into the category of special rate cut is $ 56,288.