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xxTIMURxx [149]
3 years ago
9

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

ls 15. Find the probability that a randomly selected adult has an IQ between 87 and 123.

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

0.7698

Step-by-step explanation:

If you call your random variable X, then what you are looking for is

P(87 \leq X \leq 123)

because you want the probability of X being <em>between 87 and 123.</em>

We need a table with of the normal distribution. But we can only find the table with \mu = 0 and \sigma = 1. Because of that, first we need to  <em>normalize </em>our random variable:

Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15}

(you can always normalize your variable following the same formula!)

now we can do something similar to our limits, to get a better expression:

\frac{87 - 105}{15} = \frac{-18}{15} = -1.2

\frac{123 - 105}{15} = \frac{18}{15} = 1.2

And we transform our problem to a simpler one:

P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2)

(see Figure 1)

From our table we can see that P(Z \leq 1.2) = 0.8849 (this is represented in figure 2).

Remember that the whole area below the curve is exactly 1. So we can conclude that  P(Z \geq 1.2) = 0.1151 (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then

P(Z \leq -1.2)= P(Z \geq 1.2) = 0.1151.

FINALLY:

P(Z \leq 1.2) - P(Z \leq -1.2) = 0.8849 - 0.1151 = 0.7698

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