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 $Q_3\sim 12$ and $IQR\sim 12$. 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 $\ln (\mbox{ hits }) =\ln(7.5)=2.0149$.

Problem II

A.

Because B and C are independent, it follows that

$${\bf P}(A\cup C)={\bf P}(B)+{\bf P}(C)-{\bf P}(B\cap C)={\bf P}(B)+{\bf P}(C)- {\bf P}(B)*{\bf P}(C) =.5+.4-.5*.4=.7.$$

B.

Since $B\cap A$ is a subset of $B\cup C$ we get that

$${\bf P}(B\cap C\vert B\cup C)=\frac{{\bf P}(B\cap C)}{{\bf P}(B\cup C)}=.2/.7=2/7.$$

C.

(i)

Since ${\bf P}(A\cap B)={\bf P}(B\vert A){\bf P}(A)=.6*.6$ we find that

$${\bf P}(\bar{A}\cap\bar{B})=1-{\bf P}(A\cup B)=1-[{\bf P}(A)+{\bf P}(B)-{\bf P}(A\cap B)]= 1-(.6+.5-.6*.6)=.26.$$

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

$${\bf P}(\overline{A\cap B})=1-{\bf P}(A\cap B)=1-.6*.6=.64.$$

(ii)

Since the probability that C occurs is independent of whether A or B occurs we have that

$${\bf P}(\bar{C}\vert\bar{A}\cup\bar{B})={\bf P}(\bar{C})=.6.$$

Hence,

$${\bf P}(A\cup B\cup C)=1-{\bf P}(\bar{A}\cap\bar{B}\cap\bar{C... ...f P}(\bar{C}\vert\bar{A}\cap\bar{B}){\bf P}(\bar{A}\cap\bar{B})$$

$$=1-{\bf P}(\bar{C}){\bf P}(\bar{A}\cap\bar{B})=1-.6*.26=.844.$$

Problem III

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

A.

${\bf P}(\mbox{Find on C})$

$$\begin{array}{rl} =&{\bf P}(\mbox{Find on C}\vert\mbox{Correc... ...box{Correct submenu on C is chosen})\\ =&.7*.9=.63. \end{array}$$

B.

Using previous result we find that

$$\begin{array}{rcl} {\bf P}(\mbox{Find the page})&=&{\bf P}(\m... ...sen})\\ &=&.5*1/3+.56*1/3+.63*1/3=169/300 =0.56(3). \end{array}$$

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

$$\begin{array}{rcl} {\bf P}(\mbox{The page is found})&=& {\bf ... ...& 1/3*1*.5+1/3*.9*.56+1/3*.8*.63=1508/3000=.502(6). \end{array}$$

So,

$${\bf P}(\mbox{A is chosen}\vert\mbox{The page is found})=\frac{1/3*1*.5}{1/3*1*.5+1/3*.9*.56+1/3*.8*.63}=500/1508.$$

Problem IV

A.

The probability distribution of X is shown below:

X	-1000	0	500	1000
${\bf P}$	.1	.2	.2	.5

(i)

${\bf P}(X>0)=.2+.5=.7.$

(ii)

${\bf E}(X)=-1000*.1+0*.2+500*.2+1000*.5=500.$

(iii)

${\bf Var}(X)=(-1000-500)^2*.1+(0-500)^2*.2+(500-500)^2*.2+(1000-500)^2*.5=400000$Respectively, standard deviation $\sigma=\sqrt{400000}$.

B.

(i)

${\bf P}(X>0)=\displaystyle\int_0^1(x+1)/2\,dx=x^2/4+x/2\Big\vert _0^1=3/4.$

(ii)

$${\bf E}(X)=\int_{-1}^1x(x+1)/2\,dx=x^3/6+x^2/4\Big\vert _{-1}^1=1/3.$$

C.

The distribution of daily profits, Y, is given by

Y	-1500	-500	500	1500
${\bf P}$	.1	.2	.2	.5

Hence, ${\bf E}(Y)=-1500*.1+-500*.2+500*.2+1500*.5=600.$