**Problem I**

**A.**- Since the distance from the median to lower quartile
*Q*_{1}is less than the distance from the median to the upper quartile*Q*_{3}this indicates the number of hits is skewed to the right. Also the lower whisker is smaller than the upper whisker. **B.**- Visually 36 is not above the upper whisker.
Also
*IQR*=*Q*_{3}-*Q*_{1}=19-2=17 and the upper inner fence is*Q*_{3}+1.5**IQR*=44.5. Since 36 is smaller it is not a candidate outlier. **C.**- The median is higher for recreational than for business. Since the box is bigger for recreational it also has higher variance. Business and recreational are both skewed to the right.
**D.**- Since 36 is well above the box for business it would be candidate outlier. Visually and . So limit is 12+1.5*12=30.
**E.****(i)**- The right tail would be pulled in (higher values less extreme). This would make the data look symmetric.
**(ii)**- Since logs preserve order, median of the .

**Problem II**

**A.**- Because
*B*and*C*are independent, it follows that

**B.**- Since
is a subset of
we get that

**C.****(i)**- Since
we find that

The different interpretation of this question is possible, so the full credit is granted also for the following solution:

**(ii)**- Since the probability that C occurs is independent
of whether A or B occurs we have that

Hence,

**Problem III**

(Note: Some students used the information from problem II. If the information was used correctly, no penality resulted.)

**A.**-

**B.**- Using previous result we find that

**C.**- The probability of finding the page on A is .5; for B this probability is .56; for C it is .63. So search engine C is best.
**D.**- First we note that

So,

**A.**- The probability distribution of
*X*is shown below:*X*-1000 0 500 1000 .1 .2 .2 .5 **(i)****(ii)****(iii)**- Respectively, standard deviation .
**B.****(i)****(ii)****C.**- The distribution of daily profits,
*Y*, is given by*Y*-1500 -500 500 1500 .1 .2 .2 .5