A population has a mean of µ = 80 with = 20. a. If a single score is randomly se

Question

A population has a mean of µ = 80 with = 20.

a. If a single score is randomly selected from this population, how much distance, on average, should you find between the score and the population mean?

b. If a sample of n = 4 scores is randomly selected from this population, how much distance, on average, should you find between the sample mean and the population mean?

c. If a sample of n = 100 scores is randomly selected from this population, how much distance, on average, should you find between the sample mean and the population mean?

Explanation / Answer

Answer to the question)

Given:

Mean = M = 80

Standard deviation = s = 20

.

The distance between mean and any score is expressed in terms of Number of standard deviations

This distance is calculated with the help of Z score

The score x, is converted to a Z score, which tells us how far x is from the mean in terms of standard deviation

.

Part a)

Suppose the raw score is x

Z = (x - M) / s

Z = (x - 80) / 20

.

Now suppose if assume x to be 100 , then

Z = (100- 80 ) / 20 = 1

this means 100 is 1 standard deviation above the mean

.

if x is 60, then

Z = (60 - 80) / 20 = -1

This means 60 is one standard deviation below the mean

.

Answer to part b)

if we get the sample size n , it changes the formula of Z as follows:

Z = (x - M) / (s /n)

We got n = 4

.

Thus Z = (x-80) / (20/4)

Z = (x-80) / 10

.

Now if x is 100 ,then

Z = (100 - 80) / 10 = 2

this implies 100 is 2 standard deviations above the mean

.

Part c)

if n = 100 , then

Z = (x - 80) / (20/100)

Z = (x - 80) / 2

.

Now suppose if x is 100, then

Z = (100 - 80) / 2 = 10

This imples Z is very far away from the mean . It is 10 standard deviations above the mean.

.

Note: Suppose for part a, we say the average distance is Z , then for part b the average distance is 2*Z , and for part c the average distance will be 10*Z

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