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Question Help Assume that adults have IQ scores that are normally distributed with a mean of = 100 and a standard deviation o
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Answer #1

Given that, mean (μ) = 100 and standard deviation (\sigma) = 15

We want to find, P(80 < X < 120)

\\ P(80 < X < 120) \\\\ = P(\frac {80-100}{15} < \frac {X -\mu}{\sigma} < \frac {120 -100}{15}) \\\\ = P(-1.33 < Z < 1.33) \\\\ = 2 * P(Z < 1.33) - 1 \\\\ = (2 * 0.9082) - 1 \\\\ = 1.8164 - 1 \\\\ = 0.8164

Therefore, the probabilitiy that a randomly selected adult has a IQ between 80 and 120 is 0.8164

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