Mathematics, 30.05.2020 19:03 mary9590

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.

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Answer from: gtrsoccer

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.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$-1.645 = \frac{X - 572}{51}$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 572}{51}$

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.

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