Supplementary Exercise 5.7 of IPS7e
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The number of accidents (X) has mean mu=2.2 and standard deviation sigma=1.4.
It is a discrete distribution and therefore definitely not normal (a
Poisson distribution would be one possibility).

We consider the average number of accidents during 52 weeks (Xmean, 
for simplicity of notation denoted Y).

(a) The central limit theorem (CLT) says that the average is approximately 
normally distributed with mean EY=EX=2.2 and standard deviation 
sdY=sdX/sqrt(52)=1.4/7.21=0.194.

(b) Using the approximate normal distribution, we get
  P(Y<2) = P((Y-2.2)/.194 < (2-2.2)/0.194) = P(Z<-1.03) = 0.15, 
using table or software.

(c) Fewer than 100 accidents per year corresponds to less than
100/52=1.923 accidents per week. We compute similarly
  P(Y<1.923) = P(Z<-1.43) = 0.077.
The difference to the probability in (b) is substantial although the
difference between 2 and 1.923 accidents per week seems small; relative
to the standard deviation for Y it is however quite considerable.


Minitab commands and output:
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MTB > CDF 2;
SUBC>   Normal 2.2 0.194.
Cumulative Distribution Function 

Normal with mean = 2.2 and standard deviation = 0.194

x  P( X <= x )
2     0.151287

MTB > CDF 1.923;
SUBC>   Normal 2.2 0.194.
Cumulative Distribution Function 

Normal with mean = 2.2 and standard deviation = 0.194

    x  P( X <= x )
1.923    0.0766697