Question

Suppose the length of time an iPad battery lasts can be modeled by a normal distribu-tion with meanμ= 8.2 hours and standard deviationσ= 1.2 hours.

a. What is the probability that randomly selected iPad lasts longer than 10 hours?

b. What is the probability that a randomly selected iPad lasts between 7 and 10hours?

c. What is the 3rd percentile of the battery times?

d. Suppose 16 iPads are randomly selected. What is the probability that the meanlongevity ̄Xis less than 7.9 hours?

Answer 1

Mean = = 8.2

Standard deviation = = 1.2

a) We have to find P(X > 10)

For finding this probability we have to find z score.

That is we have to find P(Z > 1.5)

P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668 ( Using z table)

So the probability that randomly selected iPad lasts longer than 10 hours is 0.0668

b) We have to find the probability that a randomly selected iPad lasts between 7 and 10hours.

That is we have to find P( 7 < X < 10)

For finding this probability we have to find z score.

That is we have to find P( - 1 < Z < 1.5)

P( - 1 < Z < 1.5) = P(Z < 1.5) - P(Z < - 1) = 0.9332 - 0.1587 = 0.7745 ( Using z table)

So the probability that a randomly selected iPad lasts between 7 and 10 hours is 0.7745

c) We have to find the 3rd percentile of the battery times.

That we have to give P(X < x) = 0.03

And we have to find the value of x.

Z value for 0.03 is - 1.88 ( Using z table)

So the 3rd percentile of the battery times is 5.944 hours.

d) n = 16

We have to find the probability that the mean longevity ̄X is less than 7.9 hours.

That is we have to find P( < 7.9)

For finding this probability we have to find z score.

That is we have to find P(Z < - 1)

P(Z < - 1) = 0.1587 ( Using z table)

So the probability that the mean longevity ̄Xis less than 7.9 hours is 0.1587