219k views
5 votes
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.

User Andrea T
by
8.5k points

1 Answer

3 votes

Answer:

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.

User Petr Felzmann
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.