The number of bicycle versus automobile accidents per week a

The number of bicycle versus automobile accidents per week at the intersection of Pardall Road and Embarcadero Del Norte varies with mean 3.2 and standard deviation 1.4. (This distribution is discrete and so it is not normal.) Let the sample mean, X, be the mean number of accidents per week at the intersection during a year (52 weeks). What is the distribution of the sample mean? What is the probability that the mean number of accidents is less than 2? What is the probability that there are on average more than 5 accidents per week?

Solution

A)

By central limit theorem:

It is approximately normally distrbuted.

It has a mean of 3.2.

It has a standard deviation of sigma/sqrt(n) = 1.4/sqrt(52) = 0.194145069.

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2      
u = mean =    3.2      
n = sample size =    52      
s = standard deviation =    1.4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -6.180945044      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -6.180945044   ) =    3.18595*10^-10 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    5      
u = mean =    3.2      
n = sample size =    52      
s = standard deviation =    1.4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    9.271417565      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   9.271417565   ) =    0 [Too close to 0]