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Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100. a) What percentage of the people taking the test score between 400 and 500

User Trevorsg
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Answer:

38.3% of the people taking the test score between 400 and 500

Explanation:

When the distribution is normal, we use 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, we have that:


\mu = 450, \sigma = 100

What percentage of the people taking the test score between 400 and 500

We have to find the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 400. So

X = 500


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


Z = (500 - 450)/(100)


Z = 0.5


Z = 0.5 has a pvalue of 0.6915

X = 400


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


Z = (400 - 450)/(100)


Z = -0.5


Z = -0.5 has a pvalue of 0.3085

0.6915 - 0.3085 = 0.383

38.3% of the people taking the test score between 400 and 500

User Bruno Shine
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