Question

1. Find the probability that a randomly selected senior high school student spends more than 26 hours but less than 29 hours.

Answer 1

Mean = 24 hours

Standard deviation = 4 hours

P(26 < x < 29) = P[(26 - 24) / 4 < z < (29 - 24) /4]

P(26 < x < 29) = (0.25 < z < 1)

Here z scores are 0.25 and 1. Now, probability is determined for each z score using normal distribution table. Probability for 0.25 is 0.60 and for 1 is 0.84. Then, the probability of a randomly selected senior high school student spends more than 26 hours but less than 29 hours is:

P(26 < x < 29) = 0.84 - 0.60 = 0.24 or 24%