Statistics 101: Midterm Exam 1

Problem I

A.
Since the distance from the median to lower quartile Q1 is less than the distance from the median to the upper quartile Q3 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=Q3-Q1=19-2=17 and the upper inner fence is Q3+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,

Problem IV
A.
The probability distribution of X is shown below:
 X -1000 0 500 1000 0.1 0.2 0.2 0.5
(i)
(ii)
(iii)
Respectively, standard deviation .
B.
(i)
(ii)
C.
The distribution of daily profits, Y, is given by
 Y -1500 -500 500 1500 0.1 0.2 0.2 0.5
Hence,