Question

The standard deviation of the sampling distribution of the sample mean is $115,000. The population standard deviation was $25,000. A random sample of 100 new home sales from this city was taken. What is the probability that the sample mean selling price was between $114,000 and $116,000?

Answer 1

0.31084

Explanation:

When given a question where by a random sample is chosen for the population, the z score formula used =

z = (x-μ)/Standard error

where x is the raw score

μ is the population mean

Standard error = σ/√n,

σ is the population standard deviation

n = number of samples

For x = $114,000, μ = $115,000, σ = 25,000, n = 100

z = (x-μ)/Standard error

= z = (x-μ)/ σ/√n

z = 114,000 - 115,000/ 25,000/√100

= -1000/25,000/10

= -1000/2500

z = -0.4

Determining the probability value from Z-Table:

P(x = 114000) = P(z = -0.4)

= 0.34458

For x = $116,000, μ = $115,000, σ = 25,000, n = 100

z = (x-μ)/Standard error

= z = (x-μ)/ σ/√n

z = 116,000 - 115,000/ 25,000/√100

= 1000/25,000/10

= 1000/2500

= 0.4

Determining the Probability value from Z-Table:

P(x = 116000) = P(z = 0.4)

= 0.65542

The probability that the sample mean selling price was between $114,000 and $116,000 is calculated as

= P(114,000< x < 116,000)

= P(-Z<x<Z)

= P(-0.4<x< 0.4

= 0.65542 - 0.34458

= 0.31084

Therefore, the probability that the sample mean selling price was between $114,000 and $116,000 is 0.31084