Question

Find the probability that a randomly selected senior high school student spends less than 21 hours or greater than 30 hours

Answer 1

Mean is 24 hours with a standard deviation of 4 hours.

Let's find the opposite (the probability that a student spends more than 21 and less than 30 hours), than subtract this probability from 1, and that will be the answer.

The probability that 21 < X < 30 is equal to the blue area under the curve.

Since $\mu$ = 24 and $\sigma =4$  we have:

P( 21 < X < 30 ) = P(21-24 < X - $\mu$ < 30 - 24) = P$(\frac{21-24}{4}<\frac{X-\mu}{\sigma} <\frac{30-24}{4} )$

Since $Z = \frac{X-\mu}{\sigma}$, $\frac{21-24}{4} = -0.75$ and $\frac{30-24}{4} = 1.5$ we have:

P( 21 < X < 30) = P(-0.75 < Z < 1.5)

Use the standard normal table to conclude that:

P(-0.75 < Z < 1.5) = 0.7066

So, P' = 1 - 0.7066 = 0.2934

Answer: P = 0.2934.

P.S. Here is the z-score table: