Download Midterm 1, Version 2, Solutions

SOLUTIONS TO MIDTERM 1
VERSION 2
1) The sorted observations are 2,3,5,20. The median is the average of the two
middle values, (3+5)/2=4. Answer is B.
2) The sorted observations are 1,9,13. The range is 13−1=12. Answer is C.
3) The sample variance is the square of the sample standard deviation,
s 2  16 2  256. Answer is C.
4) Yes, consider the data set {1,3,4}. The median is 3 and the range is 3. Answer
is A.
5) Adding 10 to all data values will add 10 to the 25th and 75th percentiles, and
will therefore have no effect on the IQR. Doubling the resulting values will
then double the IQR. Answer is A.
6) If the data points are x1 and x 2 , we get the two equations x2  x1  5,
and x1  x2  5. Solving these two equations gives x1  0, x2  5. Now, since
the sample mean is 2.5 and n−1=1, we have
s 2  (0  2.5) 2  (5  2.5) 2  6.25  6.25  12.5. Answer is B.
7) The sample space of 8 equally likely simple events is
S={HHH,HHT,HTH,HTT,TTT,TTH,THT,THH}. Note that
P(A)=P({HHH,HHT,HTH,THH})=1/2, P(B)=1/2, and
P(AB)=P({HHH,HHT,THH})=3/8. Since 3/8 is not equal to (1/2)(1/2), A
and B are not independent. So they are dependent. Answer is B.
8) Since all diamonds are red cards, we have AB=B, so that
P(AB)=P(B)=13/52=.25. Answer is A.
9) Since P(AB)=P(A)+P(B)P(AB) (general addition rule), we get P(AB)+
P(AB)=P(A)+P(B) which the problem states is less than 1. Answer is D.
10) The sample space is S={HT,TH,HH,TT}, so A , the complement of A,
consists of everything in S that is not in A. Thus, A ={HH,HT}. Answer is C.
1 n 2 1 n
xi   ( xi  x ) 2 . If we multiply this by n /( n  1) we

n i 1
n i 1
n
80 , so
get the sample variance, s 2  100. Thus, 100 
n 1
80
n 1
 .8 
 1  1 / n so 1/ n  .2 and we conclude that n=5. Answer is B.
100
n
11) We have 80 
12) If P(A)=0 then P(AB)=0, so it is not true that P(AB)>P(A). If P(A)≠0
then the independence of A and B means that P(AB)>P(A) is equivalent to
P(A)P(B)>P(A), and now dividing on both sides by P(A) we get P(B)>1,
which is impossible since probability can never exceed 1. Answer is B.
13) We have .7=P(AB)=P(A)+P(B)−P(AB)=P(A)+P(B) −P(A)P(B), so we get
.7=.5+P(B) −.5P(B), so .2=.5P(B), and we conclude that P(B)=.4. Answer is
C.
14) The economist feels that there is a probability of 1/3 that the Great Recession
will end in 2010. This corresponds to correct odds of 2:1, since odds of a:b
are correct (fair) if the probability of the event in question is b/(a+b), which
would be 1/3 in the case a=2, b=1. Answer is A.
15) If the applicant takes the GMAT n times, then Prob{At Least one score
exceeds 700}=1−Prob{No score exceeds 700}= 1  .1n . We want to find n
such that 1  .1n  .99, so .1n  .01 , so n=2. Answer is B.