For a normal population with a mean equal to 87 and a standa

For a normal population with a mean equal to 87 and a standard deviation equal to 16, determine the probability of observing a sample mean of 92 or less from a sample of size 8.

(Round to four decimal places as needed.)

Solution

Mean ( u ) =87
Standard Deviation ( sd )=16
Number ( n ) = 8
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X <= 92) = (92-87)/16/ Sqrt ( 8 )
= 5/5.6569= 0.8839
= P ( Z <0.8839) From Standard NOrmal Table
= 0.8116