Quiz #2 Problem Solving Answers
I will not provide answers for Bridget Jones/Memento Data Set. Where students went wrong is
that they put the wrong values for computational formula to find variance/sample standard
deviation. Don’t make same mistake!! Remember, you can’t have a negative variance or standard
deviation since values are squared. If an IQR is larger than another, this means that there is
greater variability in the middle 50% of the distribution;
Here are answers for Normal Distribution
i. Using column ‘B’, approx. 24% fall between 20,000 and avg in league attendance.
ii. Need to have two different z scores. Use Column ‘B’ both times add probabilities/areas; .24
+.31=.55; approx. 55% of teams fall between 20,000 and 32,000 in league attendance.
iii. Need to have two different z scores. Since both are above the mean, we have to subtract the
smaller percentage from the larger percentage; Use column ‘B’;
.2357-.1480=.0877; Approx. 9% of teams will have attendance between 28,000 and 30,000;
Xtra Credit: First, we need to find z score associated with area; Don’t just plug percentage into
formula!! Z score associated with 15%= 1.035;
25,000 + 1.035(8000)=33,280;
Interpretation: 33,280 is the value needed if an NBA team wants to be in top 15%;
Quiz #3
I.
i. Use column ‘B’; .2823
ii. Use column ‘C’; .0749
iii. Use column ‘B’; Need to have two different z scores. Since both are above the mean, we have
to subtract the smaller percentage from the larger percentage; Use column ‘B’; .4925-.4686=.03;
approx 3 % of UCB profs would get evaluations between 95 and 99;
iv. Use column ‘B’; Need to have two different z scores. Since both are below the mean, we have
to subtract the smaller percentage from the larger percentage; Use column ‘B’; .4772-.3413=.1359;
approx 14 % of UCB profs would get evaluations between 68 and 75;
II.
First, we need to find z score associated with area; Don’t just plug percentage into formula!!
i. Z score associated with 9%; Use Column ‘C’= 1.345;
77 + 1.345(9)=89;
Interpretation: 89 is the value needed if a SJSU prof wants to be in top 9%;
ii. Z score associated with 20%; Use Column ‘C’= .845;
82 + .845(7)=88; Interpretation: 88 is the value needed if a UCB prof wants to be in top
20%;
iii. Z score associated with 12%; Use Column ‘C’= 1.175;
82 + 1.175(7)=90; Interpretation: 90 is the value needed if a UCB prof wants to be in top
20%;
III.
i. First way; 25= 32 possible combinations of males and females; only one way that a women can
have five female babies in a row, and that is if they are all female; 1/32=.03;
Second way; (.5) 5; what is the probability of having a female child, AND a female child, AND
a female child, and a female child; .03;
Third way use binomial formula; p=.5; q=.5; n=5; x=5
Combination formula 5!/5(5-5)! * (.5)5 * (.5)0 = 1 * .03 * 1 =.03
Fourth way: use binomial table : p=.5; n=5; x=5; = .03
ii. p=.5; q=.5; n=20; x=8
20!/8(20-8)! * (.5)8 * (.5)12 125970 *.0039 * .00024 =.1179; approx 12% probability that they
hire exactly 8 women;
Xtra credit:
We need to use binomial formula three times. We already have answer for 8 women, now we
need to find for 6 or 7 women;
20!/8(20-8)! * (.5)8 * (.5)12 125970 *.0039 * .00024 =.1179;
20!/7(20-7)! * (.5)7 * (.5)13 77520 *.0078 * .00012 =.07;
20!/6(20-6)! * (.5)6 * (.5)14 38760 *.016 * .00006 =.04;
Approx 23% chance that they hire 6, 7, or 8 women;