Given a normal population whose mean is 615 and whose standa

Given a normal population whose mean is 615 and whose standard deviation is 27, find each of the following (use Excel to obtain more accuracy): The probability that a random sample of 4 has a mean between 617 and 622. Probability = The probability that a random sample of 16 has a mean between 617 and 622. Probability = The probability that a random sample of 26 has a mean between 617 and 622. Probability =

Solution

Mean ( u ) =615
Standard Deviation ( sd )=27
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 617) = (617-615)/27/ Sqrt ( 4 )
= 2/13.5
= 0.1481
= P ( Z <0.1481) From Standard Normal Table
= 0.55889
P(X < 622) = (622-615)/27/ Sqrt ( 4 )
= 7/13.5 = 0.5185
= P ( Z <0.5185) From Standard Normal Table
= 0.69795
P(617 < X < 622) = 0.69795-0.55889 = 0.1391                  
              
b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 617) = (617-615)/27/ Sqrt ( 16 )
= 2/6.75
= 0.2963
= P ( Z <0.2963) From Standard Normal Table
= 0.6165
P(X < 622) = (622-615)/27/ Sqrt ( 16 )
= 7/6.75 = 1.037
= P ( Z <1.037) From Standard Normal Table
= 0.85014
P(617 < X < 622) = 0.85014-0.6165 = 0.2336                  

c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 617) = (617-615)/27/ Sqrt ( 26 )
= 2/5.2951
= 0.3777
= P ( Z <0.3777) From Standard Normal Table
= 0.64718
P(X < 622) = (622-615)/27/ Sqrt ( 26 )
= 7/5.2951 = 1.322
= P ( Z <1.322) From Standard Normal Table
= 0.90691
P(617 < X < 622) = 0.90691-0.64718 = 0.2597