The yearly mileage accumulated by an automobile in a large car rental company’s fleet is normal with mean 18000 kilometres and standard deviation 1700 miles. At the end of the year the company sells 80% of its cars, keeping the 20% with the lowest mileage. Do you think a car whose year-end mileage is 17400 k.m. is likely to be kept?
This is just Z table stuff. You can play around with the values at
http://davidmlane.com/hyperstat/z_table.html
Another method.
Z = (score-mean)/SD = (17400-18000)/1700 = ?
Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability related to the Z score.
To determine whether a car with a year-end mileage of 17400 km is likely to be kept or sold, we can use the concept of Z-scores.
First, we need to calculate the Z-score for the year-end mileage of 17400 km. The Z-score measures how many standard deviations an observation is from the mean. It can be calculated using the formula:
Z = (x - μ) / σ
Where:
- Z is the Z-score
- x is the value we want to find the Z-score for (in this case, 17400 km)
- μ is the mean (18000 km)
- σ is the standard deviation (1700 km)
Calculating the Z-score for a year-end mileage of 17400 km:
Z = (17400 - 18000) / 1700
Z = -0.3529
Once we have the Z-score, we can refer to a standard normal distribution table (also known as a Z-table) to determine the probability associated with that Z-score.
The probability associated with a Z-score can be interpreted as the percentage of data points that fall below that value. In this case, we want to know the probability that a car will have a lower mileage than 17400 km (i.e., be kept).
Looking up the Z-score of -0.3529 in a Z-table, we find that the associated probability is approximately 0.3632, or about 36.32%.
Therefore, a car with a year-end mileage of 17400 km has a 36.32% chance of being kept by the car rental company. So it is likely to be kept, as the company keeps the 20% of cars with the lowest mileage.