Question

Family incomes (denoted by X) in a large city are normally distributed with mean $39 thousand and standard deviation $9 thousand. A random sample of nine incomes is taken in order to estimate the population mean income. a. Is the sampling distribution of the sample mean income ( X ) normal? Provide enough reasoning. b. What is the standard error of the sample mean income? Show your work. c. What is the probability that the sample mean income is more than $36 thousand? Show your work.

Answer 1

Solution :

Let X be a random variable which represents the family income in a large city.

Given that , X ~ N(39000, 9000)

i.e. μ = $39000 and    σ = $9000

a) If we have a normally distributed population with mean μ and standard deviation σ, then sampling distribution of sample mean follows normal distribution with mean μ and standard deviation σ/√n.

(Where, n is sample size.)

Therefore, sampling distribution of the sample mean income will be normal.

b) The standard error of sample mean is given as follows:

Where, SE is standard error, σ is population standard deviation, n is sample size.

We have, σ = $9000 and n = 9

The standard error of sample mean income is 3000.

c) We have to obtain P(x̄ > $36000).

(Where, x̄ is sample mean income of 9 incomes.)

We have,  μ = $39000, σ = $9000 and n = 9

If X ~ N(μ, σ​​​​​​2) then, x̄ ~ N(μ, σ​​​​​​2/n)

And if x̄ ~ N(μ, σ​​​​​​2/n) then

Using "pnorm" function of R we get, P(Z > -1) = 0.8413

Hence, the probability that the sample mean income is more than $36 thousand is 0.8413.

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