Question

In the journal Knowledge Quest (January/February 2002), education professors at the University of Southern California investigated children's attitudes toward reading. One study measured third through sixth graders' attitudes toward recreational reading on a 140-point scale (where higher scores indicate a more positive attitude). The mean score for this population of children was 106 points, with a standard deviation of 16.4 points. Consider a random sample of 36 children from this population, and let x represent the mean recreational reading attitude score for the sample.

(Give answers to 4 decimal places.)

a) What is ?x?

b) What is ?x?

c) What sample mean would be the cutoff for the bottom 10% of sample means. (You are being asked for the 10th percentile of sample means.)

d) Find P(x < 100).

Answer 1

a,b) x represents the general attitude of these students toward recreational reading.

c) The 10th percentile of sample means is 102.51.

d)
`P(x < 100) = 0.01390`

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this case, we have that:

The mean score for this population of children was 106 points, with a standard deviation of 16.4 points, so
`\mu = 106, \sigma = 16.4`.

a,b) What is x?

x represent the mean recreational reading attitude score for the sample. So x represents the general attitude of these students toward recreational reading.

c) What sample mean would be the cutoff for the bottom 10% of sample means. (You are being asked for the 10th percentile of sample means.)

This is the value of X when Z has a pvalue of 0.10.

Looking at the Z table, that is
`Z = -1.28`.

We are working with the mean of the sample, so we have to find the standard deviation of the sample. That is

`s = (\sigma)/(√(n)) = (16.4)/(√(36)) = 2.73`

`Z = (X - \mu)/(\sigma)`

`-1.28 = (X - 106)/(2.73)`

`X - 106 = -1.28*2.73`

`X = 102.51`

The 10th percentile of sample means is 102.51.

d) Find P(x < 100).

This is the pvalue of Z when X = 100.

`Z = (X - \mu)/(\sigma)`

`Z = (100 - 106)/(2.73)`

`Z = -2.20`

`Z = -2.20` has a pvalue of 0.01390.

So
`P(x < 100) = 0.01390`.