Answer:
0.7698
Step-by-step explanation:
If you call your random variable [tex] X [/tex], then what you are looking for is
[tex] P(87 \leq X \leq 123)[/tex]
because you want the probability of [tex] X [/tex] being between 87 and 123.
We need a table with of the normal distribution. But we can only find the table with [tex] \mu = 0 [/tex] and [tex] \sigma = 1[/tex]. Because of that, first we need to normalize our random variable:
[tex]Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15} [/tex]
(you can always normalize your variable following the same formula!)
now we can do something similar to our limits, to get a better expression:
[tex]\frac{87 - 105}{15} = \frac{-18}{15} = -1.2[/tex]
[tex] \frac{123 - 105}{15} = \frac{18}{15} = 1.2[/tex]
And we transform our problem to a simpler one:
[tex]P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2) [/tex]
(see Figure 1)
From our table we can see that [tex] P(Z \leq 1.2) = 0.8849 [/tex] (this is represented in figure 2).
Remember that the whole area below the curve is exactly 1. So we can conclude that [tex] P(Z \geq 1.2) = 0.1151 [/tex] (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then
[tex]P(Z \leq -1.2)= P(Z \geq 1.2) = 0.1151[/tex].
FINALLY:
[tex] P(Z \leq 1.2) - P(Z \leq -1.2) = 0.8849 - 0.1151 = 0.7698 [/tex]
