Question

Raw scores on standardized tests are often transformed for easier comparison. A test of reading ability has a mean of 75 and a standard deviation of 10 when given to third-graders. Sixth-graders have a mean score of 82 and a standard deviation of 11 on the same test.
What percent of sixth-graders earn a score of at least 93 on the reading test?

Answer 1

Using the normal distribution, it is found that 3.59% of sixth-graders earn a score of at least 93 on the reading test.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula, which is given by:

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

• $\mu$ is the mean.
• $\sigma$ is the standard deviation.
• It measures how many standard deviations the measure is from the mean. Each z-score has a p-value associated with it.
• 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.

• On the reading test, the mean is of 75, thus $\mu = 75$.
• The standard deviation is of 10, thus $\sigma = 10$.
• The proportion who scored above 93 is 1 subtracted by the p-value of Z when X = 93, thus:

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

$Z = \frac{93 - 75}{10}$

$Z = 1.8$

$Z = 1.8$ has a p-value of 0.9641.

1 - 0.9641 = 0.0359.

0.0359 x 100% = 3.59%

3.59% of sixth-graders earn a score of at least 93 on the reading test.

A similar problem is given at brainly.com/question/24663213