Question

Suppose that taxis in New York are driven an average of 60,000 miles per year with a standard deviation of 11,000 miles. Assume that the mileage driven for a year is normally distributed.

How many miles will the middle 95% of taxis have driven in one year?

Answer 1

From what I understand, this question is asking for the range between which the mileage can be in.
Since the question is asking for the middle 95 percent, we should be using the 97.5th percentile with the Z table to solve this question.
If I remember correctly, the formula should be P(Z<=((x bar)-(miu)/sigma)) = p
Sigma = standard deviation = 11000 miles
x bar= 60000 miles
Thus formula becomes P(Z<=((60000-(miu))/11000)=0.975
Use the Z table to find the entry where P is equal to 0.975 and set that value equal to (60000-miu)/11000, solve for miu.
Final answer for the question will be 60000 + or - miu, assuming this is what the question was asking.
I myself am studying for this course's exam right now, so I can't say for sure that the formula I remembered was correct, but this should be the general idea of how to tackle this problem.

Answer 2

Let n be the number of taxis in NY. The average distance travelled is 60,000 miles, therefore the middle 95% will have the same average as the population, the reason being the mileage is symmetrically distributed about the mean Therefore the total number of miles in one year for the middle 95% is 60,000 * 0.95 * n

Answer 3

The range of miles driven by the middle 95% can be found from the empirical rule that says:
For a normal distribution, approximately 95% of the data points lie within the range plus and minus 2 standard deviations of the population mean. In this case the range is
(60,000-22,000) to (60,000 + 22,000)

Answer 4

Thanks guys. You both helped.

Answer 5

You're welcome :)