Question

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

Answer 1

Answer:

50% of females do not satisfy that requirement

Step-by-step 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:

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

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 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

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

$Z = \frac{900 - 900}{200}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

50% of females do not satisfy that requirement