1. ## tricky statistics question

the average for licnesed drivers in the country is an average of 42.6 yrs with a standard deviation of 12. The distribution is approx... normal.
a. a researcher obtained a sample of n=36 drivers who received parking tickets. The average age for these drivers was m= 40.7. Is this a reasonable outcome for a sample of n=36 or is this sample mean an extreme value?
b. The same researcher also obtained a sample of n-16. drivers who received speeding tickets. The average for this sample was M=34.4. Is this a reasonable outcome for a sample of n=16 or is the sample mean very different from what would usually be expected? (compute the z-score for the sample mean)

2. Originally Posted by sf415415
the average for licnesed drivers in the country is an average of 42.6 yrs with a standard deviation of 12. The distribution is approx... normal.
a. a researcher obtained a sample of n=36 drivers who received parking tickets. The average age for these drivers was m= 40.7. Is this a reasonable outcome for a sample of n=36 or is this sample mean an extreme value?
b. The same researcher also obtained a sample of n-16. drivers who received speeding tickets. The average for this sample was M=34.4. Is this a reasonable outcome for a sample of n=16 or is the sample mean very different from what would usually be expected? (compute the z-score for the sample mean)
Let Y be the random variable sample mean.

Using a well known result about the sampling distribution of the mean:

Y ~ Normal $\left(\mu = 42.6, ~ \sigma = \frac{12}{\sqrt{n}}\right)$.

(a) Calculate $\Pr(Y \leq 40.7)$

(b) Calculate $\Pr(Y \leq 34.4)$

Calculation of z-scores is done using the formula $Z = \frac{Y - \mu}{\sigma}$.