Question

Suppose the mean of the daily use of an iPad among all US iPad owners is 5 hours and the standard deviation is 3 hours. The distribution of daily usage is known to be NOT normal. We select 36 observations at random. What is the probability that a daily usage average of 36 US iPad owners is above 8 hours?

P(X¯>8)=P(Z>8−35/36−−√)

	P(X¯>8)=P(Z>8−53/36−−√)

	P(X¯>8)=P(Z>8−35)

Suppose the mean of the daily use of an iPad among all US iPad owners is 5 hours and the standard deviation is 3 hours. The distribution of usage is known to be NOT normal. We select 25 observations at random. We use a normal approximation for the sample mean?

	True
	False

If X ~ N(200, 10), which of the following is the 95th percentile?

	X(95th) = 10 + 1.645*200
	X(95th) = 200 + 1.96*10
	X(95th) = 200 + 1.645*10
	X(95th) = 1.645

If X ~ N(200, 10) which of the following is true?

	Z = (X - 200)/10 ~ N(0,1)
	Z = (X - 200^2)/10 ~ N(0,1)
	Z = (X - 200)/10^2 ~ N(0,1)

Answer 1

1) The sampling distribution of the sample mean is approximately normally distributed with sample mean and sample standard deviation

Where ,

Now ,

; From standard normal distribution table

2) True

Yes , for large n , we use the normal approximation for sample mean.

3)

The 95th percentile is ,

............(I)

From standard normal distribution table , .............(II)

From (I) and (II) , we get ,

4)