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The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 900 and a standard deviation of 200. If a college includes a minimum score of 900 among its requirements, what percentage of females do not satisfy that requirement

User Alpants
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1 Answer

6 votes

Answer:

50% of females do not satisfy that requirement

Explanation:

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 problem, we have that:


\mu = 900, \sigma = 200

If a college includes a minimum score of 900 among its requirements, what percentage of females do not satisfy that requirement

This is scores lower than 900, which is the pvalue of Z when X = 900.

So


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


Z = (900 - 900)/(200)


Z = 0


Z = 0 has a pvalue of 0.5

50% of females do not satisfy that requirement

User Nabbit
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