1The number of accidents per week at a hazardous intersectio

1.The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. This distribution takes only whole-number values, so it is certainly not Normal. Let x¯ be the mean number of accidents per week at the intersection during a year (52 weeks). Consider the 52 weeks to be a random sample of weeks. What is the mean of the sampling distribution of x¯?

2.Referring to question 1, what is the standard deviation of the sampling distribution of x¯?

3.Referring to question 1, why is the shape of the sampling distrbution of x¯ approximately Normal?

4.Referring to question 1, what is the approximate probability that x¯ is less than 2?

Solution

1.

By central limit theorem, the mean of the sampling distirbution is the population mean,

u(X) = 2.2 [ANSWER]

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2.

By central limit theorem, the standard deviation of the sampling distribution is given by

sigma(X) = sigma/sqrt(n) = 1.4/sqrt(52) = 0.194145069 [ANSWER]

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3.

Because of the central limit theorem.

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4.

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 =    2.2      
n = sample size =    52      
s = standard deviation =    1.4      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.030157507      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.030157507   ) =    0.151468037 [ANSWER]