Download 21) Consider a random variable X with the following

CHAPTER 7 REVIEW
21) Consider a random variable X with the following probability distribution:
X
20
21
22
23
P(X)
0.2
0.3
0.2
0.1
a) Find P(X ≤ 22)
b) Find P(X > 21)
c) Find P(21 ≤ X < 24)
d) Find P(X ≤ 21 or X > 23)
ANS:
a) Find P(X ≤ 22) = 0.2 + 0.3 + 0.2 = 0.7
b) Find P(X > 21) = 0.2 + 0.1 + 0.2 = 0.5
c) Find P(21 ≤ X < 24) = 0.3 + 0.2 + 0.1 = 0.6
d) Find P(X ≤ 21 or X > 23) = 0.2 + 0.3 + 0.2 = 0.7
24
0.2
22) In the casino game of roulette, a ball is rolled around the rim of the circular bowl while a wheel
containing 38 slots into which the ball can drop is spun in the opposite direction from the rolling ball; 18
of the slots are red, 18 are black, and 2 are green. A player bets a set amount, say $1, and wins $1 if the
ball falls into the color slot the player has wagered on. Assume a player decides to bet that the ball will
fall into one of the red slots.
a) What is the probability that the player will win?
b) What is the expected return on a single bet of $1 on red?
ANS:
a) 18 of the 38 slots are winners, so P(win if bet on red) = 18/38 = 0.474
b) The probability distribution for this game is
Outcome
Win
Lose
X
1
-1
P(X)
18/38
20/38
 18 
 20 
E(X) =  x  1   (1)   0.052 . The player will lose 5.2¢ for each dollar bet.
 38 
 38 
23) A radio station is running a lottery to raise money for a local charity. The prizes are $10, $50, $100,
and a grand prize of $1000. The chances of winning these amounts are 0.25, 0.15, 0.09, and 0.01
respectively. What are your total expected winnings (minus costs) if you pay $1 for a ticket?
ANS:
 x   xp( x)  1(0.5)  9(0.25)  49(0.15)  99(0.09)  999(0.01)  $28
24) The scores for the top three golfers on a high school golf team are used to determine which high
schools advance to the regional level. The Central High team’s top three players have mean scores and
standard deviations of:
Player 1
Player 2
Player 3
μ
89.5
94.4
97.2
σ
2.3
4.5
3.9
What are the mean score and standard deviation for the Central High team?
ANS:
Each of the high school player’s scores are random variables, and the player’s scores are
independent from one another. The mean of the sum of random variables is the sum of the means of the
individual random variables.  x  y   x   y . So,  x  y  z   x   y   z  89.5  94.4  97.2  281.1.
For the standard deviation, we work first with the variance, since the variance of the sum is the sum of the
 x2 y   x2   y2 . For the golfers that would be
 x2 y  z   x2   y2   z2  (2.3) 2  (4.5) 2  (3.9) 2  40.75, and thus  x  y  z   x2 y  z  40.75  6.38.
25) The grading at Point Loma High gives a B for grades between 86 and 93. On the English final for
seniors, what proportion of the class would get a B if the grades were normally distributed with a mean
grade of 86.34 and standard deviation of 14.23?
ANS:
P(86 ≤ X ≤ 93) = normalcdf(86, 93, 86.34, 14.23) = 0.1896
26) The mean GPA for Point Loma High is 2.9, with the standard deviation of 0.5. Assuming the GPAs
are normally distributed, what GPA score will place a student in the top 5% of the class?
ANS:
invNorm(0.95, 2.9, 0.5) = 3.72
27) Webb is a baseball fanatic. He keeps his own statistics on the major league teams and individual
players. For the 350 regular starters, Webb has found their mean batting average is 0.229, with a standard
deviation of 0.024. His sister is appalled that baseball players get paid the salaries they do and get a hit
less than 25% of their attempts at bat. To further her argument, she asks for the following information:
a) What proportion of players hit more than 25% of the times they are at bat?
b) Since the players with the top ten batting averages get cash bonuses, what is the lowest batting average
is that will receive a bonus?
ANS:
a) P(X > 0.25) = normalcdf(0.25, 1E99, 0.229, 0.024) = 0.1922
b) The top ten players would represent 10/350 = 0.02857, or approximately 2.86% of the players.
invNorm((1-0.02857), 0.299, 0.024) = 0.2746. Specifically, this says that the top 10 players have batting
averages of 0.2746 or higher.
28) Sophia was recently promoted to assistant manager at a small women’s clothing store. One of her
duties is to fill out order forms for women’s shirts, which come in sizes 6, 7, 8, 9, 10, 11, and 12. She
would like to determine how many shirts of each size to order. At first, she thought of ordering exactly
the same number of shirts from each of the available sizes, but then she decided against doing that,
because there might be a greater demand for certain sizes than for others. She looked up sales receipts
from the past three months and summarized the information as follows:
Shirt Size
6
7
8
9
10
11
12
Number Sold
85
122
138
154
177
133
92
a) Prepare a probability distribution of the number of shirts sold for each size.
b) What is the probability that a randomly selected customer will request a shirt of size at least 11?
c) Compute the expected shirt size of a random shopper and the standard deviation of the shirt size.
d) If Sophia plans to order a total of 1000 shirts, how many shirts of size 8 should she order?
ANS:
a) The total number of shirts sold is 901. Using this information, we can compute the probability of each
shirt size being sold.
Shirt Size
P(x)
6
85/901 = 0.09
7
122/901 = 0.14
8
138/901 = 0.15
9
154/901 = 0.17
10
177/901 = 0.20
11
133/901 = 0.15
12
92/901 = 0.10
b) P(a customer will request a shirt of size at least 11) = 0.15 + 0.10 = 0.25
c) The expected shirt size:
E ( x)    6(0.09)  7(0.14)  8(0.15)  9(0.17)  10(0.20)  11(0.15)  12(0.10)  9.09
The standard deviation of shirt size:
  (6  9.09) 2 (0.09)  (7  9.09) 2 (0.14)  ...  (12  9.09) 2 (0.10)  1.84
d) Suppose Sophia plans to order a total of 1000 shirts:
From the probability distribution, P(Shirt size = 8) = 0.15
Therefore, she needs to order 1000(0.15) = 150 shirts of size 8.
29) A company markets 16-ounce bottles of jam. The mean amount of jam per bottle is 16 ounces, with a
standard deviation of 0.1 ounces. The mean weight of the glass bottles holding the jam is 5 ounces, with a
standard deviation of 0.5 ounces.
a) What is the mean weight of a filled bottle?
b) What is the standard deviation of the weight of a filled bottle?
c) When shipped to stores, 12 bottles are packed randomly in each box. What is the mean and standard
deviation of the weights of these random groupings of 12 bottles?
d) The mean weight of the empty boxes is 50 ounces, with a standard deviation of 4 ounces. What is the
mean weight and the standard deviation of the weights of the filled boxes?
e) If the amount of jam per bottle, the weight of the bottles (with their lids), and the weight of the boxes
are approximately normally distributed, what percent of boxes will weigh more than 320 ounces?
ANS:
Let A = the amount of jam per bottle
Let B = the weight of the bottles (with lids)
Let C = the weight of a box packed with 12 bottles
It is known that μA = 16 ounces, σA = 0.1 ounces, and
μB = 5 ounces, σB = 0.5 ounces.
a) Let W = weight of a filled bottle = weight of bottle content + weight of a bottle (with its lids)
Therefore, the mean weight of a filled bottle is
W   A   B  16  5  21 ounces
b) The standard deviation of the weight of a filled bottle is
 W   A2   B2  0.12  0.5 2  0.5099  51 ounces
c) Twelve bottles are grouped together randomly. Let G = total weight of a group of 12 bottles.
The mean weight of the group  G  12W  12W  12(21)  252 ounces
The standard deviation of the weight of the group is 12 2  W2  12 2 (0.51) 2  6.12 ounces
d) Let E = the weight of an empty box and C = the weight of a packed box = G + E
The mean weight of a packed box is  C   G   E  252  50  302 ounces.
The standard deviation of the weights of the packed boxes is
 C   G2   E2  6.12 2  4 2  7.31 ounces.
e) Because all weights are approximately normally distributed, N(302, 7.31).
Therefore, P(C > 320) = normalcdf(320, 1E99, 302, 7.31) = 0.0069
Thus, 0.69% boxes will weigh more than 320 ounces.
30) In a game of chance at a school carnival, a player rolls a dodecahedral die—a die with 12 faces. The
faces are numbered from 1 through 12. On the basis of the roll of the die, the player wins a prize. In the
table below, the monetary values of the prizes are listed with their corresponding die rolls.
Die Roll
1, 2, 3, 4, or 5
6, 7, 8, or 9
10 or 11
12
Prize Value
$0.50
$1.00
$2.00
$4.50
a) What are the mean and standard deviation of the value of prizes won by a player who plays this game?
b) It costs $1.00 to play the game. What is the net gain/loss in monetary value from the player’s
perspective?
ANS:
a) The mean of the probability distribution is
5
4
2
1
 x  (0.50)   (1.00)   (2.00)   (4.50)   $1.25
 12 
 12 
 12 
 12 
The standard deviation is
5
4
2
1
 X2  (0.5  1.25) 2    (1.00  1.25) 2    (2.00  1.25) 2    (4.50  1.25) 2    1.2292
 12 
 12 
 12 
 12 
 x  1.2292  $1.1087
b) If the player can expect to win a prize worth $1.25 on each play, and the player must pay $1.00 to play
the game, then the net gain/loss from the player’s perspective is a gain of $0.25 per game.
31) A brand of model rocket kit, when finished, produces rockets with a mean mass of 134 grams and a
standard deviation of 4.0 grams. The model rocket engine required to fly the rocket is purchased
separately and has a mass with mean 36 grams and standard deviation 2.0 grams. If a randomly selected
engine is inserted into a randomly selected rocket, what are the mean and standard deviation of the total
mass?
ANS:
Since the rocket and engine are both randomly selected, their weights are independent of each other.
 Rocket  Engine   Rocket   Engine  134  36  170 grams
2
2
2
2
2
20  4.47 grams
 Rocket
 Engine   Rocket   Engine  4.0  2.0  20, so  
32) The heights of 18-year-old men are approximately normally distributed with mean 68 inches and
standard deviation 3 inches. What is the probability that an 18-year-old man selected at random is less
than 71 inches tall?
ANS: 0.8413