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5.13 IQ scores on the WAIS test approximate a normal curve
with a mean of 100 and a standard deviation of 15. What IQ
score is identified with
a. the upper 2 percent, that is, 2 percent to the right (and 98
percent to the left)?
b. the lower 10 percent?
c. the upper 60 percent?
d. the middle 95 percent? [Remember, the middle 95 percent
straddles the line perpendicular to the mean (or the 50th
percentile), with half of 95 percent, or 47.5 percent, above
this line and the remaining 47.5 percent below this line.]
e. the middle 99 percent?

Respuesta :

Using the normal distribution, it is found that the IQ scores identified with the desired percentages are:

  • a) 130.81.
  • b) 80.8
  • c) 96.2
  • d) [70.6, 129.4]
  • e) [61.37, 138.63]

Normal Probability Distribution

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 100, hence [tex]\mu = 100[/tex].
  • The standard deviation is 15, hence [tex]\sigma = 15[/tex].

Item a:

This is the 98th percentile, which is X when Z has a p-value of 0.98, so X when Z = 2.054.

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

[tex]2.054 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = 2.054(15)[/tex]

[tex]X = 130.81[/tex]

Item b:

This is the 10th percentile, which is X when Z has a p-value of 0.1, so X when Z = -1.28.

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

[tex]-1.28 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = -1.28(15)[/tex]

[tex]X = 80.8[/tex]

Item c:

This is the 100 - 60 = 40th percentile, which is X when Z has a p-value of 0.4, so X when Z = -0.253.

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

[tex]-0.253 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = -0.253(15)[/tex]

[tex]X = 96.2[/tex]

Item d:

This is between the 2.5th percentile(X when Z = -1.96) and the 97.5th percentile(X when Z = 1.96), hence:

Z = -1.96:

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

[tex]-1.96 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = -1.96(15)[/tex]

[tex]X = 70.6[/tex]

Z = 1.96:

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

[tex]1.96 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = 1.96(15)[/tex]

[tex]X = 129.4[/tex]

Item e:

This is between the 0.5th percentile(X when Z = -2.575) and the 99.5th percentile(X when Z = 2.575), hence:

Z = -2.575:

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

[tex]-2.575 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = -2.575(15)[/tex]

[tex]X = 61.37[/tex]

Z = 2.575:

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

[tex]2.575 = \frac{X - 100}{15}[/tex]

[tex]X - 100 = 2.575(15)[/tex]

[tex]X = 138.63[/tex]

You can learn more about the normal distribution at https://brainly.com/question/24663213

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