# Stat 330 Sample Solution Homework 8 1 Central Limit

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```					Stat 330                               Sample Solution                                   Homework 8
1    Central Limit Theorem
A bank accepts rolls of pennies and gives 50 cents credit to a customer without counting the contents.
Assume that a roll contains 49 pennies 30 percent of the time, 50 pennies 60 percent of the time, and 51
pennies 10 percent of the time.
(a) Find the expected value and the variance for the amount that the bank loses on a typical roll.
Let X be the money in cents that the bank loses on one roll. Then P (X = 1) = 0.3, P (X = 0) = 0.6,
and P (X = −1) = 0.1.
That yields an expected loss for the bank of
E[X] = 0.3 + 0 − 0.1 = 0.2
and a variance of
V ar[X] = (1 − 0.2)2 · 0.3 + (0 − 0.2)2 · 0.6 + (−1 − .2)2 · 0.1 = 0, 36

(b) Estimate the probability that the bank will lose more than 25 cents in 100 rolls.
Let Z be the loss on 100 rolls, then Z = X1 + X2 + ... + X100 , where Xi is the loss on roll i. All Xi
are independent and have the same distribution.
According to the Central Limit Theorem Z then have a normal distribution with mean 100 · 0.2 = 20
and variance 100 · 0.36 = 36.
Then
P (Z > 25) = 1 − P (Z ≤ 25) = 1 − N20,36 (25) = 1 − N0,1 (5/6)

(c) Estimate the probability that the bank will lose exactly 25 cents in 100 rolls.
Using the normal approximation, P (Z = 25) ≈ 0 because Z is approximately normally distributed, and
the probability that a continuous random variable is exactly equal to a number is zero.
Another acceptable solution is this:

P (Z = 25) = P (Z < 25.5) − P (Z ≤ 24.5) = N20,36 (25.5) − N20,36 (24.5)

(d) Estimate the probability that the bank will lose any money in 100 rolls.

P (Z ≥ 0) = 1 − N20,36 (0) = N0,1 (20/6) ≈ 0.9996

(e) How many rolls does the bank need to collect to have a 99 percent chance of a net loss?
Let Zn be the loss of the bank on n rolls, then
√
P (Zn ≥ 0) = 1 − N0.2n,0.36n (0) = N0,1 (0.2/0.6 n) ≥ 0.99.
√
If 0.2/0.6 n ≥ 2.33 the probability of a net loss is at least 0.99. Therefore
√
n ≥ 3 · 2.33
and n ≥ 48.9, i.e. the bank has to collect 50 rolls to have a 99% probability of a net loss.

1
2    Central Limit Theorem - again
A true-false examination has 48 questions. June has probability 3/4 of answering a question correctly. April
just guesses on each question. A passing score is 30 or more correct answers. Compare the probability that
June passes the exam with the probability that April passes it. June’s score has distribution B48,.75 , so the
probability that June’s score is 30 or more is 1 − B48,.75 (29) = 0.9627. In case your calculator doesn’t give
an answer, you will have to use a normal approximation to the Binomial distribution (based on the Central
Limit Theorem):

29 − 36
B48,.75 (29) ≈ N48·.75,48·.75·.25 (29) = N36,9 (29) = N0,1     √       = N0,1 (−2.33) = 0.0264,
9
which results in an overall probability of 0.9736.
April’s score has distribution B48,.5 , so the probability that April’s score is 30 or more is 1 − B48,.5 (29) =
0.0297, or based on the approximation

12 − 29
P (A ≥ 29) ≈ 1 − N12,9 (29) = 1 − N0,1                  = 7.6e − 09.
3

In any case, June has a much better chance of passing the exam than April.

3    Central Limit Theorem - and again
A rookie is brought to a baseball club on the assumption that he will have a .300 batting average. (Batting
average is the ratio of the number of hits to the number of times at bat.) In the ﬁrst year, he comes to bat
300 times and his batting average is .267. Assume that his at bats can be considered Bernoulli trials with
probability .3 for success. Could such a low average be considered just bad luck or should he be sent back
to the minor leagues?
To get a batting average of 0.267 the rookie had 80 hits in the 300 times he was at bats. Let X be the number
of times he hits - X has a Binomial distribution with n = 300 and p = 0.3.
The probability to get a batting average of 0.267 or less is then the probability to have 80 hits or less in 300
times at bats:

P (X ≤ 80) = B300,0.3 (80)

we won’t ﬁnd this probability in a table, since n is so large. Instead, we are using a normal approximation.
B300,0.3 ≈ N300·0.3,300·0.3·0.7 . Then:
80 − 90
P (X ≤ 80)    = B300,0.3 (80) ≈ N90,63 (80) = N0,1 ( √      )=
63
= N0,1 (−1.26) = 1 − N0,1 (1.26) = 1 − 0.9131 = 0.0869.
The rookie therefore has a probability of approx. 9% to show a batting average of 0.267, even though his true
standard is 0.3. Whether this is enough reason to send him back is a diﬀerent question - but if the probability
was even lower ( less than 5% or less than 1%, maybe), the trainer should send him back to the minors.

4    Central Limit Theorem - yet again
A restaurant feeds 400 customers per day. On the average 20 percent of the customers order apple pie.
(a) Give a range for the number of pieces of apple pie ordered on a given day such that you can be 95
percent sure that the actual number will fall in this range.

2
Each customer can be considered a Bernoulli trial, denoted by X.

E[X] = .2
2
σX = .2(.8) = .16
If we add 400 of these together, we can use the central limit theorem and assume that it will be
approximately normal:

E[Y ] = 400E[X] = 80
2       2
σY = 400σX = 64

σY =      2
σY = 8

Since 95% is conveniently two standard deviations from the mean, our maximum distance from the
mean is 16 pies, giving a range of 64–96 pies.
(b) How many customers must the restaurant have, on the average, to be at least 95 percent sure that the
number of customers ordering pie on that day falls in the 19 to 21 percent range?
Given n, customers, we have the following the the distribution of the total number of pies:

E[Z] = nE[X] = .2n
2     2
σZ = nσX = .16n
2
√
σZ = σZ = .4 n
Since we want the fraction, not the total, we divide by n:
Z
E[ ] = .2
n
.16
σ2 =
Z
n     n
2   .4
σZ =     σZ = √
n
Since 95% is two standard deviations, we need to solve as follows:
.4
2 √ = .01
n
√
.8 = .01 n
√
80 = n
6400 = n
The restaurant must have 6400 customers to have a 95% conﬁdence that the number of pies ordered
will be between 19% and 21%.

3

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