Question

the weekly earnings of all families in a large city have a normal probability distribution with mean of $950 and standard deviation of $95.

a) if someone is randomly selected what's the probability that his/her weekly weekly earning will be less than $1000.

b) if you you select a random sample of 25 families from the city. what is the mean, the standard deviation, and the shape of the sampling distribution of the sample mean?

c) what's the probability that the mean (average) weekly earnings of the sample will be less than $1000.

Answer 1

Solution :

Let X be a random variable which represents the weekly earnings of all families in a large city.

Given that,

Mean (μ) = $950

SD (σ) = $95

a) We have to obtain P(X < $1000)

We know that, if X ~ N(μ, σ²) then

Using "pnorm" function of R we get, P(Z < 0.5263) = 0.7007

Hence, if someone is randomly selected the probability that his/her weekly earning will be less than $1000 is 0.7007.

b) If we draw a random sample of size n from a normally distributed population which has mean μ and standard deviation σ, then the sampling distribution of sample mean (x̄) follows normal distribution with mean μ and standard deviation σ/√n. And sampling distribution of sample mean is bell shaped.

i.e. If X ~ N(μ, σ²) then, x̄ ~ N(μ, σ²/n)

We have, μ = $950,  σ = $95 and n = 25

The mean of the sampling distribution of sample mean is, μ = $950.

The standard deviation of the sampling distribution of sample is, .

The sampling distribution of sample mean will be bell shaped.

c) We have to obtain P(x̄ < 1000).

We know that, if X ~ N(μ, σ²) then, x̄ ~ N(μ, σ²/n).

And if x̄ ~ N(μ, σ²/n) then,

Using "pnorm" function of R we get, P(Z < 2.6316) = 0.9958

Hence, the probability that the mean (average) weekly earnings of the sample will be less than $1000 is 0.9958.

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