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Time spent using​ e-mail per session is normally​ distributed, with mu equals μ=12 minutes and sigma...

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?

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Expert Solution

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)


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