AP STATISTICS
Quiz-6.2 Transforming and Combining Random Variables
Last 4 of Student ID____________
Period______
1. The manager of a children’s puppet theatre has determined that the number of adult tickets he sells for a Saturday afternoon
show is a random variable with a mean of 28.3 tickets and a standard deviation of 5.3 tickets. The mean number of children’s
tickets he sells is 42.5, with a standard deviation of 8.1.
(a) The adult tickets sell for $10. Let A = the money he collects from adult tickets on a random Saturday. What are the mean and
standard deviation of A?
µ10A=10(28.3)=$283 𝝈10A=10(5.3)=$53
(b) The children’s tickets sell for $6. Let T = the money he collects from all ticket sales (adults and children) on a random
Saturday. Assume (unrealistically, perhaps) that the number of tickets sold to adults is independent of the number sold to
children. What are the mean and standard deviation of T?
µ10A+6C=10(28.3)+6(42.5)=$538 𝝈10A+6C=√(𝟏𝟎 ⋅ 𝟓. 𝟑)𝟐 + (𝟔 ⋅ 𝟖. 𝟏)𝟐 =$71.91
(c) It costs $300 for the manager to put on each puppet show. Let P = the profit from a random Saturday’s show. What are the
mean and standard deviation of P?
µ10A+6C-300=538-300=$238 𝝈10A+6C-300=$71.91 (standard deviation is not affected by subtraction)
2. A game show host has developed a new game called “Grab All You Can." Here’s how it works: a contestant reaches his dominant
hand (i.e. right hand for right-handed people) into a jar of $10 bills and grabs as many as he can in one handful. Then he does the
same thing with his non-dominant hand in a jar of $20 bills. Research with many volunteers has determined that the mean
number of $10 bills drawn is 68 with a standard deviation of 9.5, and the mean number of $20 dollar bills is 58, with a standard
deviation of 7.8.
(a) If D = the amount of money, in dollars, that a randomly-selected contestant grabs from the “$10 grab,” find the mean and
standard deviation of D.
µ10D=10(68)=$680 𝝈10D=10(9.5)=$95
(b) If T = the total amount of money, in dollars, that a contestant grabs from both jars, find the mean and standard deviation of T.
µ10D+20N=10(68)+20(58)=$1840 𝝈10D+20N =√(𝟏𝟎 ⋅ 𝟗. 𝟓)𝟐 + (𝟐𝟎 ⋅ 𝟕. 𝟖)𝟐 =$182.65
(c) The game’s rules are that a contestant must pay $500 (from previous winnings) for one round of play (that is, one grab from
each jar). If G = how much the contestant gains from one round of play, find the mean and standard deviation of G.
µ10D+20N-500=1840-500=$1340 𝝈10D+20N-500=$182.65 (standard deviation is not affected by subtraction)
3. Suppose that the mean height of policemen is 70 inches with a standard deviation of 3 inches. And suppose that the mean
height for policewomen is 65 inches with a standard deviation of 2.5 inches. If heights of policemen and policewomen are
normally distributed, find the probability that a randomly selected policewoman is taller than a randomly selected policeman.
Let D=difference in height between random male and random female
µD=70-65=5 𝝈D=√𝟑𝟐 + 𝟐. 𝟓𝟐 =3.91
P(D<0) normalcdf(-100,0,5,3.91)=0.1004
4. Mr. Voss and Mr. Cull bowl every Tuesday night. Over the past few years, Mr. Voss’s scores have been approximately normally
distributed with a mean of 212 and a standard deviation of 31. During the same period, Mr. Cull’s scores have also been
approximately normally distributed with a mean of 230 and a standard deviation of 40. Assuming their scores are independent,
what is the probability that Mr. Voss scores higher than Mr. Cull on a randomly-selected Tuesday night?
Let D=difference in scores between Mr. Cull and Mr. Voss
µD=230-212=18 𝝈D=√𝟒𝟎𝟐 + 𝟑𝟏𝟐 =50.61
P(D<0) normalcdf(-100,0,18,50.61)=0.3512