Question

The scores on one portion of a standardized test are approximately Normally distributed, N(572, 51). a. Use the 68-95-99.7 rule to estimate the range of scores that includes the middle 95% of these test scores. b. Use technology to estimate the range of scores that includes the middle 90% of these test scores.

Answer 1

a) The range of scores that includes the middle 95% of these test scores is between 470 and 674.

b) The range of scores that includes the middle 90% of these test scores is between 488.1 and 655.9.

Explanation:

68-95-99.7 rule:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Z-score:

Problems of normally distributed samples are 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)`

The Z-score measures how many standard deviations the measure is from the mean. 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 get the probability that the value of the measure is greater than X.

In this question:

Mean
`\mu = 572`, standard deviation
`\sigma = 51`

a. Use the 68-95-99.7 rule to estimate the range of scores that includes the middle 95% of these test scores.

By the 68-95-99.7 rule, within 2 standard deviations of the mean.

572 - 2*51 = 470

572 + 2*51 = 674

The range of scores that includes the middle 95% of these test scores is between 470 and 674.

b. Use technology to estimate the range of scores that includes the middle 90% of these test scores.

Using the z-score formula.

Between these following percentiles:

50 - (90/2) = 5th percentile

50 + (90/2) = 95th percentile.

5th percentile.

X when Z has a pvalue of 0.05. So when X when Z = -1.645.

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

`-1.645 = (X - 572)/(51)`

`X - 572 = -1.645*51`

`X = 488.1`

95th percentile.

X when Z has a pvalue of 0.95. So when X when Z = 1.645.

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

`1.645 = (X - 572)/(51)`

`X - 572 = 1.645*51`

`X = 655.9`

The range of scores that includes the middle 90% of these test scores is between 488.1 and 655.9.