Answer to Question #189334 in Statistics and Probability for ABHISHEK SINGH

Question

Let ܺ denote the number of scores on a test. If ܺ is normally

distributed with mean 100 and variance 225, find the probability

that ܺ does not exceed 130.

Answer 1

$\sigma$ = $\sqrt{\sigma^2}$ = $\sqrt{225}$ =15

The probability that X < 30 is equal to the blue area under the curve.

Since $\mu=100$ and $\sigma=15$ we have:

P(X<130) = P(X-$\mu$ < 130-100) = P$(\frac{X-\mu}{\sigma}<\frac{130-100}{15})$

Since $\frac{X-\mu}{\sigma} = Z$ and $\frac{130-100}{15} = 2$  we have:

P(X<130) = P(Z<2)

Use the standard normal table

 to conclude that:

P(Z<2) = 0.9772

Answer: P = 0.9772.