In: Math
Time spent using e-mail per session is normally distributed, with mu equals μ=12 minutes and sigma equals σ=3 minutes. Assume that the time spent per session is normally distributed. Complete parts (a) through (d). b) If you select a random sample of 50 sessions, what is the probability that the sample mean is between 11.5 and 12 minutes?
µ = 12
σ = 3
n= 50
we need to calculate probability for ,
11.5 ≤ X ≤ 12
X1 = 11.5 , X2 =
12
Z1 = (X1 - µ )/(σ/√n) = -1.18
Z2 = (X2 - µ )/(σ/√n) = 0.00
P ( 11.5 < X <
12 ) = P ( -1.18
< Z < 0.00 )
= P ( Z < 0.00 ) - P ( Z
< -1.18 ) =
0.5000 - 0.1193 =
0.3807
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excel formula for probability under Z value is
=normsdist(z)