Name:
Section _____________________________________________________________
Student number:
STAT 101-002
Statistics 101, Homework 7
Due Tuesday, April 20, 2010, 6pm
Instructions:
This assignment is to be completed individually. You may discuss this homework with
the other students in this course (any section), but not with students who are not currently
enrolled in Statistics 101. You may also ask the instructors and the TAs if you need help,
but you must attempt the questions and submit your written solution by yourself.
Although you may discuss the problems with each other, you should not read each
others solutions. Do not copy verbatim. If you use material from outside this class, it
must be referenced in your solution. Honor code applies to these instructions.
In writing your solutions, make them complete. You should show your work and explain
your conclusions clearly and concisely. Your solutions should be typed by editing this
file and inserting your text and JMP output where needed. Do not forget to fill in your
name and section above.
Submit your solutions before the deadline by handing in a printed paper copy AND by emailing the edited MS Word file in an attachment to [e-mail].
Readings: Chapter 12 and Chapter 16.
INSTRUCTIONS: Use proper subscripts in your notation for random variables. In MS
Word, you can toggle in and out of subscript mode by hitting ctrl-=. (Superscripts are
done with ctrl-+, but you wont need them.) Also, use the Greek letters and for the
E() and SD() of i.i.d. r.v.s; finally, use the symbol for approximately equal to. If you
cant find them on your computer, copy them from here to your solutions.
1. The fraction of M&Ms that are blue is 20%. A bag of M&Ms contains 56 random
M&Ms. Answer the following questions, stating clearly the random variables and the
assumptions used when applying the CLT.
a) What is approximately the chance that more than half of the M&Ms in the bag are
blue? Proceed in steps: Define suitable random variables (2pts). State the task in
terms of a probability statement (1pt). Check the applicability of the CLT (2pts).
Do the calculations by collecting all the ingredients, stating the general formulas
first and then filling in the numbers (4pts). When calculating the probability (1pt)
state exactly at what step you are using the CLT and mark that step with the symbol instead of = (1pt). Finally express the probability more intuitively as
about one in bags are more than half blue (1pt).
YOUR SOLUTION:
1. Binary random variable b: 1 if the M&M is blue; 0 if any other color is chosen.
2. B= b +b ++b and signifies the number of blue M&Ms. P(B>28).
1
2
n
3. CLT is valid since n=56>20. Assumptions: the sums and averages of the
random variables are normally distributed, independence, and identical
distribution.
4. N=56 and p=0.2
E(b ) = p=.2 and Var(b ) = p(1 p)=.16.
i
i
E(B) = 56*E(b ) = 56(0.2) = 11.2.
i
Var(B) = 56*Var(b ) = 56(0.2)(0.8) = 8.96.
i
1/2
=8.96 =2.99.
P(B > 28) P (z > (28 11.2)/2.99) P(z > 16.8/2.99) P(z > 5.6) 1.0768-8.
About one in 92,870,000 bags is majority blue.
b) What is the chance that fewer than 5 of the M&Ms are blue? To solve this problem draw on part a) (3pts). Again state the result in the form about one in
(1pt).
YOUR SOLUTION:
E(B) = 56*E(b ) = 56(0.2) = 11.2
i
1/2
=8.96 =2.99
P(B < 5) P( z < (5 11.2)/2.99) P(z< - 2.074) less than 0.02275.
About one in 52 bags has fewer than 5 blue M&Ms.
2. In a lottery game in a Casino one can win 0, 1 or 2 dollars with equal probabilities. A
ticket to participate in the game costs $1.10. In one evening, 514 people participate in
the game. What is the approximate probability that the Casino will lose? Proceed in
small steps, explain the finances, and state the assumptions you are making. As usual,
express the final result as once in (11pts).
YOUR SOLUTION:
a. Equal probabilities: P(Xi=0)=1/3, P(Xi=1)=1/3, P(Xi=2)=1/3.
b. Player profits (considering cost and probabilities): P(Pi=-1.10)=1/3,
P(Pi=-.10)=1/3, P(Pi=.90)=1/3
c. Total profits of players: Pt=P1+P2++P514
d. Casino loses if players make any profits: P(casino loses)=P(Pt>0)
e. E(Pi)= 1/3*(-1.1+-.1+.9)=-0.10
E(Pt)=514*E(Pi)=-51.4
Var(Pt) = N * Var(Pi) = 514 * (1/3 * (-1.10 + 0.1)^2 + 1/3 * (-0.1 + 0.1)^2 + 1/3 *
(0.9 + 0.1)^2) = 342.66
SD(Pt)=(342.66)1/2 18.5
Z=(-51.4-0)/18.5=2.78
f. P(Pt>0)=P(Z>2.78)=.9972608
3. The weight of people that fly between San Francisco and Seattle is 160 lb on average,
with a standard deviation of 30 lb.
a) What is the chance that the total weight 36 of people on such a flight will be
above 6000lb? State your assumptions and criticize them if they seem unrealistic.
(5pts)
YOUR SOLUTION:
i = 160; and i=30
Wt=W1++W36
t= 36* i=5760
Var(Wt)=36*302=32400; and =180
Z=(5760-6000)/180= -1.333
P(Wt>6000)=P(Z>-1.34) .9
Assumption: Equal distribution of weights which may be unrealistic given the
group of people. Perhaps it was a flight in which all the passengers are models, or
it could be a group of sumo wrestlers.
b) What is the probability that the average weight of the 36 people on the flight is
less than 150 lb? (4pts)
YOUR SOLUTION:
I am defining average as: A=(W /36) = [(w + w + + w )/36].
(A) = (Wt/36)=(1/36)* (W ) = 160
T
t
1
2
36
Var(A) = (1/1296)*Var(Wt); Var(Wt)= 36*i2=32400
Var(A)=25 and = 5 since it is the square root of the variance.
CLT dictates: P(A < 150) P(z < (150 160)/5) P(z < -2) 0.025
4. A tire manufacturer warranties its tires to last at least 20,000 miles or you get a new
set of tires. In its experience, a set of these tires lasts on average 25,000 miles with
SD 4,000 miles. Assume that the wear is normally distributed. The manufacturer
profits $200 on each sold set, and replacing a set costs it $400.
a) What is the probability that a set of tires wears out before 20,000 miles? Use the
r.v. X = miles till wear-out. Round the result to just one decimal. (Note: This
problem has nothing to do with the CLT, only with the stated assumption.) (3pts)
YOUR SOLUTION:
(X) = 25,000 and (X) = 4,000.
Assumption: Random variable is has normal prob distribution.
P(x < 20,000) = P(z < (20,000 25,000)/4,000) = P (z < -1.25) = 0.1.
b) What is the probability that the manufacturer turns a profit on selling a set to one
customer? Again round to just one decimal.
(2pts)
YOUR SOLUTION:
Profit means the company does not have to pay anyone for faulty tires so P(profit on
one tire) = P( x 20,000) = 1 P(x < 20,000) = 0.9.
c) If the manufacturer sells 500 sets of tires, what is the probability that its net profit
exceeds $75,000? Assume that the purchases are made around the country and
that the drivers experience independent amounts of wear. Start by defining a r.v.
that describes the net revenue from one customer at a time. (Use the rounded
probabilities.) (10pts)
YOUR SOLUTION:
The profit on tire, P = 200 with probability 0.9 (from b) = 200 with probability
0.1
Using the definition of expected value, E(P ) = 200(0.9) 200(0.1) = 160.
Using the definition of variance Var(P ) = (200 160) (0.9) + (200 160) (0.1) =
14400.
and = 120.
P =p+p ++p .
E(Pi)*500= E(P ) = 500*160= 80,000
Var(P ) = 500*14400 = 7,200,000
SE(P ) = 2683.3.
CLT: P (P > 75,000) P(z > (75,000 80,000)/2683.3) 0.97.
i
i
i
T
1
2
500
T
T
T
T
2
2
5. Hoping to lure more shoppers downtown, a city builds a new public parking garage in
the central business district. The city plans to pay for the structure through parking
fees. The consultant who advised the city on this project predicted that parking revenues would average $1,300 per day, with a standard deviation of $150.
a) Assuming the consultant is correct, what is the chance that in a two-month period
(44 weekdays) more than $55,000 will be collected? State the necessary assumptions to justify your calculations. (6pts)
YOUR SOLUTION:
(X) = 1,300 and (X) = 150; n = 44.
The money collected over that period of time is defined as S = x + + x .
E(S)=44*E(X) = 44(1,300) = 57,200
Var(S) = 44*(150 ) = 44(22,500) = 990,000
(S) = 995.
CLT: P(S > 55,000) P(z > (55,000 57,200)/995) = P(z > -2.2) = 1 P(z <
-2.2) = 0.99.
Assumptions: The sums are independent. They are also identically
distributed.
1
44
2
b) During the two-month period (44 weekdays), daily fees collected averaged
$1,264. Based on this information, how should you interpret the average $1,264?
Is it the true but unknown expected value of daily parking revenue? If yes, why?
If not, what is it? (2pts)
YOUR SOLUTION:
The $1264 is not an expected value, but rather a sample average.
c) What is the standard error for the mean of revenue from 44 days (2pts)? What
does the mean $1,264 say about the consultants assumptions? Put differently, if
the consultants assumptions were true, would $1,264 be extreme? (2pts)
YOUR SOLUTION:
I define sample average as A = (x + x + + x )/44.
SE(A) = Var(A) = Var(x )/n = 150 /44 = 22.6.
The Z test results: (1264 1300)/22.6 = -1.59.
The Sample average is about 2 SD away from the 1300 provided mean. This
would not be an extreme situation.
1
i
2
44
2