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A grading scale is set up for 1000 students' test scores. It assumes the
scores are normally distributed with a mean score of 75 and a standard
deviation of 15.
How many students will score between 48 and 75?
hint: recall how to get the standard deviation. What does that mean in
this context? How could you use that to understand how many test
scores could be in one standard deviation?
464

Sagot :

Answer:

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

Step-by-step explanation:

Normal Probability Distribution:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

It assumes the scores are normally distributed with a mean score of 75 and a standard deviation of 15.

This means that [tex]\mu = 75, \sigma = 15[/tex]

How many students will score between 48 and 75?

First we find the proportion, which is the pvalue of Z when X = 75 subtracted by the pvalue of Z when X = 48. So

X = 75

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{75 - 75}{15}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5

X = 48

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{48 - 75}{15}[/tex]

[tex]Z = -1.8[/tex]

[tex]Z = -1.8[/tex] has a p-value of 0.0359

1 - 0.0359 = 0.4641

Out of 1000:

0.4641*1000 = 464

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

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