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2 votes
IQ scores are normally distributed with a mean of 100 and a

standard deviation of 15. What percentage of people have an
IQ score between 82 and 98, to the nearest tenth?

User JeffB
by
7.9k points

1 Answer

3 votes

Answer:

33.2%

Explanation:

Calculate the z-scores.

z = (x − μ) / σ

z₁ = (82 − 100) / 15

z₁ = -1.2

z₂ = (98 − 100) / 15

z₂ = -0.133

Use a chart or calculate to find the probability.

P(-1.2 < Z < -0.133) = P(Z < -0.133) − P(Z < -1.2)

P(-1.2 < Z < -0.133) = 0.4470 − 0.1151

P(-1.2 < Z < -0.133) = 0.3319

Rounded to the nearest tenth, the percentage is 33.2%.

User Vivek Mohan
by
7.6k points

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