The Intelligence Test Scale in a school is composed of a number of subtests. On one subtest, the raw scores have a mean of 35 and a standard deviation of 6. Assuming these raw scores form a normal distribution: a) What number represents the 65th percentile (what number separates the lower 65% of the distribution)? b) What number represents the 90th percentile? c) What is the probability of getting a raw score between 28 and 38? d) What is the probability of getting a raw score between 41 and 44?.

Answers

Answer 1

Using the normal distribution, it is found that:

a) The 65th percentile is of 37.31.

b) The 90th percentile is of 42.68.

c) There is a 0.5705 = 57.05% probability of getting a raw score between 28 and 38.

d) There is a 0.0919 = 9.19% probability of getting a raw score between 41 and 44.

In a normal distribution 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]

It 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, which is the percentile of X.

In this problem:

The mean is of 35, hence [tex]\mu = 35[/tex]The standard deviation is of 6, hence [tex]\sigma = 6[/tex]

Question a:

The 65th percentile is X when Z has a p-value of 0.65, so X when Z = 0.385.

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

[tex]0.385 = \frac{X - 35}{6}[/tex]

[tex]X - 35 = 0.385(6)[/tex]

[tex]X = 37.31[/tex]

The 65th percentile is of 37.31.

Question b:

The 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28.

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

[tex]1.28 = \frac{X - 35}{6}[/tex]

[tex]X - 35 = 1.28(6)[/tex]

[tex]X = 42.68[/tex]

The 90th percentile is of 42.68.

Question c:

The probability is the p-value of Z when X = 38 subtracted by the p-value of Z when X = 28, hence:

X = 38:

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

[tex]Z = \frac{38 - 35}{6}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a p-value of 0.6915.

X = 28:

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

[tex]Z = \frac{28 - 35}{6}[/tex]

[tex]Z = -1.17[/tex]

[tex]Z = -1.17[/tex] has a p-value of 0.121.

0.6915 - 0.121 = 0.5705.

There is a 0.5705 = 57.05% probability of getting a raw score between 28 and 38.

Question d:

The probability is the p-value of Z when X = 44 subtracted by the p-value of Z when X = 41, hence:

X = 44:

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

[tex]Z = \frac{44 - 35}{6}[/tex]

[tex]Z = 1.5[/tex]

[tex]Z = 1.5[/tex] has a p-value of 0.9332.

X = 41:

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

[tex]Z = \frac{41 - 35}{6}[/tex]

[tex]Z = 1[/tex]

[tex]Z = 1[/tex] has a p-value of 0.8413.

0.9332 - 0.8413 = 0.0919

There is a 0.0919 = 9.19% probability of getting a raw score between 41 and 44.

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


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Answers

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