Time spent using email per session is normally distributed with u = 8 minutes and o = 2 minutes.?

Z= X-Mu/standard deviation/sqrt(n) a) P(7.8 ≤ sample mean ≤ 8.2) = P ( (7.8-8)/(2/sqrt(25)) ≤ Z ≤ (8.2-8)/(2/sqrt(25)) ) P(-.2/.4 ≤ Z ≤ .2/.4) = P(-.5 ≤ Z ≤ .5) From the Z table for Standard Normal Curve Areas you get .6915 - .3085 = .3830 b) P(7.5 ≤ sample mean ≤ 8) = P((7.5-8)/(2/sqrt(25)) ≤ Z ≤ (8-8)/(2/sqrt(25)) ) = P(-.3/.4 ≤ Z ≤ 0) = P(-.75 ≤ Z ≤ 0) Once again from the Z table for Standard Normal Curve Areas you get .5000- .2266 = .2734 c) n= 100 instead of 25 so the calculation is P(7.8 ≤ sample mean ≤ 8.2) = P( (7.8-8)/(2/sqrt(100)) ≤ Z ≤ (8.2-8.0)/(2/sqrt(100)) ) = P( -.2/.2 ≤ Z ≤ .2/.2) = P( -1 ≤ Z ≤ 1) From the Z table for Standard Normal Curve Areas you get .8413 - .1587 = .6826.

Here’s a hint that should get you along: The probability of a sample of a normal random variable of size 1 falling between a given standard deviation of the mean can be gotten by transforming to the standard normal curve: Z = (X – u)/o and then looking at your table you got. That method holds for EVERY normal random variable X. Well, if X is normal with mean you and standard deviation o, then it so happens that the average of a sample of size n is also normal, with mean still u, but with standard deviation o/sqrt(n).

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