Z= X-Mu/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
Source(s):
My own experience of over 9 years as a high school and college math tutor.

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