Find the probability that a randomly selected senior high school student spends less than 21 hours or greater than 30 hours
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 = 24 and we have:
P( 21 < X < 30 ) = P(21-24 < X - < 30 - 24) = P
Since , and 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:
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