Urban planners will use electronic traffic counters to count the number of vehic

Question

Urban planners will use electronic traffic counters to count the number of vehicles that travel on a specific road. This information can be used to identify roads that need to be improved or expanded based on traffic needs. Suppose that, from historical studies, Ducan Road in Kent Country averages 2,165 vehicles per day with a standard deviation of 485 vehicles per day. A traffic counter was used on this road on 33 days that were randomly selected. a) What is the probability that the sample mean is less than 2,000 vehicles per day? b) what is the probanility that the sample mean is more than 2,200 vehicles per day? c) what is the probability that the sample mean is between 2,100 ans 2,300 vehicles per day? d) suppose the sample mean was 2,400 vehicles per day. Does this result support the stated population mean for this road?

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2000      
u = mean =    2165      
n = sample size =    33      
s = standard deviation =    485      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.954335746      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.954335746   ) =    0.025330762 [ANSWER]

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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 =    2200      
u = mean =    2165      
n = sample size =    33      
s = standard deviation =    485      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.414556067      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.414556067   ) =    0.339233457 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    2100      
x2 = upper bound =    2300      
u = mean =    2165      
n = sample size =    33      
s = standard deviation =    485      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.769889839      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.599001974      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.220682621      
P(z < z2) =    0.945089918      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.724407297   [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    2400      
u = mean =    2165      
n = sample size =    33      
s = standard deviation =    485      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.78344788      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.78344788   ) =    0.002689225

As this is a very small probability, then NO, THIS DOES NOT SUPPORT THE STATED POPULATION MEAN FOR THIS ROAD. [ANSWER]  

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