Time spent using e-mail per session is normally distributed with mean of 8 minutes and standard deviation of 2 minutes. If you select a random sample of 25 sessions
a)What is the probability that the sample mean is between 7.8 and 8.2 minutes?
b)What is the probability that the sample mean is between 7.5 and 8 minutes
c)If you select a random sample of 100 sessions what is the probability that the sample mean is between 7.8 and 8.2 minutes
2007-11-12 10:05:14 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
These are all done in essentially the same way. I will solve (a).
Key fact: (xbar - µ)/(σ/√n) has standard normal distribution,
where xbar is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size.
P(7.8 < xbar < 8.2) = P( (7.8 - 8.0)/(2/√25) < (xbar - 8.0)/(2/√25) < (8.2 - 8.0)/(2/√25) )
= P(-0.5 < (xbar - 8.0)/(2/√25) < 0.5)
= N_z(0.5) - N_z(-0.5)
= 0.6915 - 0.3085 = 0.3830
The other two are solved very similarly. You should get 0.3944 for (b) and 0.6826 for (c)
2007-11-12 10:49:33 · answer #1 · answered by Ron W 7 · 0⤊ 0⤋