Question

The nicotine content in a single cigarette of a particular brand is a random variable having a probability distribution with a mean of 4.9 milligrams (mgs) and a standard deviation of 0.2 mg. n = 100 of these cigarettes are chosen at random and the nicotine content of each cigarette is measured.
What is the probability that the mean nicotine content of this sample is between 4.87 mgs to 4.96 mgs?

Answer 1

µ =    4.9                              
σ =    0.2                              
n=   100                              
we need to calculate probability for ,                                  
4.87   ≤ X ≤    4.96                          
X1 =    4.87   ,   X2 =   4.96                  
                                  
Z1 =   (X1 - µ )/(σ/√n) =   -1.50                          
Z2 =   (X2 - µ )/(σ/√n) =   3.00                          
                                  
P (   4.87   < X <    4.96   ) =    P (    -1.50   < Z <    3.00   )
                                  
= P ( Z <    3.00   ) - P ( Z <   -1.50   ) =    0.9987   -    0.0668   =    0.9318

[ excel formula for probability from z score is =NORMSDIST(Z) ]

the probability that the mean nicotine content of this sample is between 4.87 mgs to 4.96 mgs = 0.9318