Question

In: Statistics and Probability

According to one study, an average student spends 4.4 hours on leisure activities per day. Assuming...

According to one study, an average student spends 4.4 hours on leisure activities per day. Assuming that the standard deviation of leisure activities per day is .5 hour. You took a sample of 16 students. Calculate the following:

a. What is the probability that sample mean of leisure activities per day is more than 4.5 hours?

b. Sample mean of leisure activities per day lies between 4.6 and 4.9 hours?

c. Sample mean of leisure activities per day is less than 4 hours?

Expert Solution

µ = 4.4

sd = 0.5

n = 16

a)

= P(Z > 0.8)

= 0.2119

b)

= P(1.6 < Z < 4)

= P(Z < 4) - P(Z < 1.6)

= 1 - 0.9452

= 0.0548

c)

= P(Z < -3.2)

= 0.0007

orchestra answered 2 years ago

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