Download Solutions to Problem Set #6 Section 5.1 10.(Paraphrased) The US

Solutions to Problem Set #6
Section 5.1
10.(Paraphrased) The US Census wants to estimate the number of people not counted by the
census. In a given locality, let N be actual the number of people who live there and let n1 be
the number of people counted by the first census and let n2 be the number of people who were
counted by the second census. Let n12 be the number of people who were counted both times.
(a) Given N, n1 and n2 , let X denote the number of people counted both times. Find the
probability that X = k, where k is a fixed positive integer between 0 and n2 .
h(N, n1 , n2 , k) =
n1
k
N −n1
n2 −k
N
n2
(Alternatively, h(N, n2 , n1 , k) will also work).
(b) Now assume that X = n12 . Find the value of N which maximizes the expression in
part (a). Hint: Consider the ratio of the expressions for successive values of N .
Fixing k = n12 , we can find the maximum for N by seeing when
positive to negative. This occurs when N = nn112n2 .
h(N +1,n1 ,n2 ,n12 )
h(N,n1 ,n2 ,n12 )
goes from
16. Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson approximation to estimate the
probability that the operator will miss at most one call if she takes a 5-minute coffee break.
Let X be the number of calls that are missed. If we assume that X is Poisson, then
“success” is defined to be “missing a phone call.” Thus, we have n = 5 minutes = 5 · 602
seconds and p = .01, so λ = np = 5 · 602 · .01 = 180 seconds = 3 minutes. Thus, the
−3 k
distribution is P (X = k) = e k!3 . Therefore, the probability of missing at most one call is:
4
e−3 · 31 e−3 · 30
+
= 3.
1!
0!
e
18. A baker blends 600 raisins and 400 chocolate chips into a dough mix and, from this,
makes 500 cookies.
(a) Find the probability that a randomly picked cookie will have no raisins.
Let X be the number of raisins in the cookie. If we assume that X has a Poisson
distribution, then “success” is defined by the probability that a raisin will end up in a given
1
600
cookie. Then, p = 500
and n = 600, so λX = n · p = 500
. Thus, P (X = 0) = e−λX = 0.3012.
(b) Find the probability that a randomly picked cookie will have exactly two chocolate
chips.
Let Y be the number of chocolate chips in the cookie. If we assume that Y has a
Poisson distribution, then “success” is defined by the probability that a chocolate chip will
1
Solutions to Problem Set #6
end up in a given cookie. Then, p =
P (Y = 2) =
e−λY λ2Y
2!
1
500
and n = 400, so λY = n · p =
400
.
500
Thus,
= 0.1437.
(c) Find the probability that a randomly chosen cookie will have at least two bits (raisins
or chips) in it.
P (X + Y = 2) = P (X = 0, Y = 2) + P (X = 1, Y = 1) + P (x = 2, Y = 0) = 0.595.
28. An airline finds that 4 percent of the passengers that make reservations on a particular
flight will not show up. Consequently, their policy is to sell 100 reserved seats on a plane
that only has 98 seats. Find the probability that every person who shows up for the flight will
find a seat available.
Let X be the number of ticket-holding passengers who do not show up. Then, if X ≥ 2,
everyone who does show up will find a seat. So, we want to calculate
P (X ≥ 2) = 1 − P (X = 0) − P (X = 1).
The probabilities on the right-hand side can be calculated using either the binomial or
Poisson distribution. With Poisson, we have λ = np = 100 ∗ .04 = 4, so the right-hand side
of the equation is 1 − e−λ − λe−λ = 1 − 5e−4 .
38. A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random
sample of 5 items is chosen to be inspected. Find the probability that the sample contains
exactly one defective item.
(a) if the sampling is done with replacement
Let X represent the number of defective whips. Then, since we’re sampling with replacement, X is a binomial random
with “success” defined to mean “being defective.”
5 variable
5
15 4
1
Therefore, P (X = 1) = 1 · ( 20 ) · ( 20 ) = 0.3955.
(b) if the sampling is done without replacement
If the sampling is done without replacement, then our distribution for X is hypergeometric. Thus, we just use the formula
5 20−5
P (X = 1) = h(20, 5, 5, 1) =
2
1
5−1
20
5
= 0.4402.