AP STATISTICS
Chapter 7 Review
Formulae: E(X) = m X = å xi pi
Var(X) = s X2 = å (xi - m X )2 pi
1. When drilling for oil, there are three possible outcomes. The first is a dry well, resulting in no
income, a wet well, resulting in an income of $400,000, and a gusher, resulting in an income of
$1,500,000. The probability distribution shows the frequencies for each outcome with X
representing the income from the oil well. Find the mean of X. m X = $310,000
X
P(X)
0
.5
400,000
.4
1,500,000
.1
2. If the cost of drilling is $200,000 give the mean of the random variable for the profit from the
drilling operation above. m X -200,000 = $110,000
3.
2
4
The figure above is a density curve of the random variable X. Find P(1.2 £ X £ 3.3) .
1
(3.3 -1.2)æ ö = .525
è 4ø
4. In a particular game, a fair die is tossed. If the number of pips is either 3 or 5 you win $2, if
number of pips is 1 you win $5, and if the number of pips is an even number you win nothing. Let
X be the amount that you win. Find the expected value of X.
X
0
2
5
P(X)
3/6
2/6
1/6
E(X) = m X = $1.50
Reference for #5,6: The weight of medium-sized tomatoes selected at random from a large bin at
the local supermarket is a random variable with mean m = 190 grams and standard deviation s = 10
grams.
5. Suppose we pick four tomatoes from the bin at random and put them in a bag. The weight of the
bag is a random variable with a mean (in grams) of
mX+X+X+X = mX + mX + mX + mX = 4 mX = 4(190) = 760g
6. Suppose we pick four tomatoes from the bin at random and put them in a bag. The weight of the
bag is a random variable with a standard deviation (in grams) of
2
2
2
2
2
2
s X+
X+X+ X = s X + s X + s X + s X = 4s X = 4(100)
s X+ X+X+ X = 400 = 20g
Reference for #7 through 10: The following birds are regulars at backyard feeders in North
Carolina. Let X represent the weight of a Carolina Chickadee and Y equal the weight of the Northern
1
Cardinal. The mean, m X = 10g , and the standard deviation, s X = 1.5g for Carolina Chickadees
compare with the mean mY = 40g , and the standard deviation, s Y = 5g for Northern Cardinals.
7. Find the mean weight of a new variable Z defined as the combined weights of two Carolina
Chickadees. mZ = mX+X = mX + mX = 2(10g) = 20g
8. Find the standard deviation of the weight of Z, the combined weights of two Carolina Chickadees.
s X2 + X = s X2 + s X2 = 2(1.5)2
s X+ X = 2.1213g
9. Find the mean weight of two Northern Cardinals and one Carolina Chickadee.
mY +Y +X = mY + mY + mX = 40 + 40 +10 = 90g
10. Find the standard deviation of the weight of two Northern Cardinals and one Carolina Chickadee.
s Y2 +Y +X = s Y2 + s Y2 + s X2 = 25 + 25 + 2.25 = 52.25
s Y +Y +X = 52.25 = 7.2284g
11. Flying squirrels overwinter in communal groups in eastern North America. A random sample of
such sites produced the following distribution of data, where X is the number of individuals at one
site. (This data is fabricated.)
X
P(X)
1
.10
2
.1
3
.29
4
.3
6
.2
7
.01
a) Given a randomly selected flying squirrel overwintering site, what is the probability that there
will be more than 3 flying squirrels overwintering? P(X > 3) = .3+ .2 + .01 = .51
b) What is the probability that 4 or fewer flying squirrels will overwinter in a given site?
P(X £ 4) = .1+ .1+ .29 + .3 = .79
c) What is P(X ³ 2) ? P(X ³ 2) = 1- P(X < 2) = 1- .1 = .9
12. Let X and Y represent independent random variables with means, m X = 4 and mY = 6 , and standard
deviations, s X = 1.5 and s Y = 2.0 , respectively. Let Z = X - Y .
a) Calculate m Z . mZ = mX-Y = mX - mY = 4 - 6 = -2
b) Calculate s Z .
s Z2 = s X2 -Y = s X2 + s Y2 = 1.5 2 + 22 = 6.25
s Z = 6.25 = 2.5
c) How would the mean and standard deviation of a new random variable W = X + Y differ from Z
in terms of mean and standard deviation. mW increases to 10 while s W = 2.5 , the same value
as for s X
d) Calculate m3+ X . m3+X = 3+ m X = 3+ 4 = 7
e) Calculate m 7 X . m7 X = 7 m X = 7 ( 4 ) = 28
f) Calculate s 3+ X . s 3+X = s X = 1.5
g) Calculate s 7 X . s 7 X = 7s X = 7(1.5) = 10.5 or
2
s 27 X = 49s 2X = 49(1.5)2
s 7 X = 10.5