Question

assume the adults have IQ scores that are normally distributed with a mean of 100.2 and a standard deviation of 17.7. find the probability that a randomly selected adult has an IQ greater than 128.9. The probability that randomly selected adult from this group has an IQ greater than 128.9 is ____.

Answer 1

ANSWER:

Given that,

assume the adults have IQ scores that are normally distributed with a mean of 100.2 and a standard deviation of 17.7. find the probability that a randomly selected adult has an IQ greater than 128.9. The probability that randomly selected adult from this group has an IQ greater than 128.9 is ____.

mean = = 100.2

standard deviation = = 17.7

Z = (x-)/ = (x-100.2)/17.7

We need to find

P(x>128.9)

Now,

P(x>128.9) = P((x-)/ > (128.9-100.2)/17.7)

P(x>128.9) = P(Z > 28.7/17.7)

P(x>128.9) = P(Z > 1.62)

P(x>128.9) = 1-P(Z < 1.62)

P(x>128.9) = 1-0.94738 (From z score table)

P(x>128.9) = 0.05262

The probability that randomly selected adult from this group has an IQ greater than 128.9 is 0.05262.

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