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Intelligence quotients​ (IQs) measured on the Stanford Revision of the Binet Simon Intelligence Scale are normally distributed with a mean of 100 and a standard deviation of 16. Determine the percentage of people who have an IQ between 115 and 140.

Answer :

ajeigbeibraheem

Answer:

the percentage of people who have an IQ between 115 and 140 is 16.79%

Step-by-step explanation:

From the information given:

We are to  determine the percentage of people who have an IQ between 115 and 140.

i.e

P(115 < X < 140) = P( X  ≤ 140) - P( X  ≤ 115)

[tex]P(115 < X < 140) = P( \dfrac{X-100}{\sigma}\leq \dfrac{140-100}{16})-P( \dfrac{X-100}{\sigma}\leq \dfrac{115-100}{16})[/tex]

[tex]P(115 < X < 140) = P( Z\leq \dfrac{140-100}{16})-P( Z\leq \dfrac{115-100}{16})[/tex]

[tex]P(115 < X < 140) = P( Z\leq \dfrac{40}{16})-P( Z\leq \dfrac{15}{16})[/tex]

[tex]P(115 < X < 140) = P( Z\leq 2.5)-P( Z\leq 0.9375)[/tex]

[tex]P(115 < X < 140) = P( Z\leq 2.5)-P( Z\leq 0.938)[/tex]

From Z tables :

[tex]P(115 < X < 140) = 0.9938-0.8259[/tex]

[tex]P(115 < X < 140) = 0.1679[/tex]

Thus; we can conclude that the percentage of people who have an IQ between 115 and 140 is 16.79%

Using the normal distribution, it is found that 82.02% of people who have an IQ between 115 and 140.

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

The proportion of people who have an IQ between 115 and 140 is the p-value of Z when X = 140 subtracted by the p-value of Z when X = 115, hence:

X = 140:

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

[tex]Z = \frac{140 - 100}{16}[/tex]

[tex]Z = 2.5[/tex]

[tex]Z = 2.5[/tex] has a p-value of 0.9938.

X = 115:

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

[tex]Z = \frac{115 - 100}{16}[/tex]

[tex]Z = -0.94[/tex]

[tex]Z = -0.94[/tex] has a p-value of 0.1736.

0.9938 - 0.1736 = 0.8202.

0.8202 = 82.02% of people who have an IQ between 115 and 140.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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