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The scores of students on a standardized test are normally distributed with a mean of 300 and a standarddeviation of 40.

(a) What proportion of scores lie between 220 and 380 points?
(b) What is the probability that a randomly chosen student scores is below 260?
(c) What percent of scores are above 326.8 points?

User Manube
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Answer:

a) 0.9544 = 95.44% of scores lie between 220 and 380 points.

b) 0.1587 = 15.87% probability that a randomly chosen student scores is below 260.

c) 25.14% of scores are above 326.8 points.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 300 and a standard deviation of 40.

This means that
\mu = 300, \sigma = 40

(a) What proportion of scores lie between 220 and 380 points?

This is the p-value of Z when X = 380 subtracted by the p-value of Z when X = 220.

X = 380


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


Z = (380 - 300)/(40)


Z = 2


Z = 2 has a p-value of 0.9772.

X = 220


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


Z = (220 - 300)/(40)


Z = -2


Z = -2 has a p-value of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% of scores lie between 220 and 380 points.

(b) What is the probability that a randomly chosen student scores is below 260?

This is the p-value of Z when X = 260. So


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


Z = (260 - 300)/(40)


Z = -1


Z = -1 has a p-value of 0.1587.

0.1587 = 15.87% probability that a randomly chosen student scores is below 260.

(c) What percent of scores are above 326.8 points?

The proportion is 1 subtracted by the p-value of Z when X = 326.8. So


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


Z = (326.8 - 300)/(40)


Z = 0.67


Z = 0.67 has a p-value of 0.7486.

1 - 0.7486 = 0.2514

0.2514*100% = 25.14%

25.14% of scores are above 326.8 points.

User Hyomin Kim
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