Answer:
a).
standard deviation of sample means = sd/sqrt(n) = 25/sqrt(100) =2.5
z value for 273, z =(273-273)/2.5 =0
z value for 275, z =(275-273)/2.5 =0.8
P( 273< mean x < 275) = P( 0<z<0.8)
=P( z < 0.8) – P( z < 0)
=0.7881 -0.5000
=0.2881
Excel function used to probability
P( z <0.8): =NORM.S.DIST(0.8,TRUE)
P( z <0): =NORM.S.DIST(0,TRUE)
b).
standard deviation of sample means = sd/sqrt(n) = 3.5/sqrt(49) =0.5
z value for 76, z =(76-75)/0.5 =2
P( mean x > 76) = P( z > 2)
= 0.0228
Excel function: P( z >2): =1-NORM.S.DIST(2,TRUE)
c).
standard deviation of sample means = sd/sqrt(n) = 2.8/sqrt(64) =0.35
z value for 70.8, z =(70.8-69.8)/0.35 =2.86
P( mean x > 70.8) = P( z > 2.86)
=0.0021
Excel function: P( z >2.86): =1-NORM.S.DIST(2.86,TRUE)
hapter 8: Sampling Distribution 1. The distribution of i is normal: n 2 30. 2. Be...
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