The nicotine content in a single cigarette of a particular b

The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 mg and standard deviation 0.1 mg. If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0.88? (Round your answers to four decimal places.)

P(x < 0.88) =

P(x < 0.87) =

Solution

Mean ( u ) =0.9
Standard Deviation ( sd )=0.1
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
P(X < 0.88) = (0.88-0.9)/0.1/ Sqrt ( 100 )
= -0.02/0.01= -2
= P ( Z <-2) From Standard NOrmal Table
= 0.0228                  
P(X < 0.87) = (0.87-0.9)/0.1/ Sqrt ( 100 )
= -0.03/0.01= -3
= P ( Z <-3) From Standard NOrmal Table
= 0.0013