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Use probability distribution models
to solve straightforward problems
exam
ial
Chapter 6
The binomial distribution
notes
Essent
Covers AS 90646 (Statistics and Modelling 3.6)
Scholarship
Statistics and
Modelling
1. The game of Crown and Anchor is played using three identical fair six-sided dice. The symbols on the
faces are a heart, diamond, club, spade, crown and anchor. A gambler pays $1 as a stake to the casino
to play this game, and bets on one of the symbols – eg crowns, coming up. The dice are then tossed
and the casino pays out as follows:
3 crowns
2 crowns
1 crown
0 crowns
Casino pays $3 and returns the gambler’s stake
Casino pays $2 and returns the gambler’s stake
Casino pays $1 and returns the gambler’s stake
No payment
a. Write down the name for the distribution that gives the number of crowns that appear and write
down its parameter(s).
b. Complete this table for the distribution of the gain to the casino:
X
1
–1
–2
–3
5
P(X = x)
72
c. Calculate the expected gain to the casino if the gambler plays the game of Crown and Anchor
60 times.
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Ans. p. 9
2 Scholarship Statistics and Modelling (Chapter 6)
d. Suggest how the payments could be adjusted to make the game fair. Make sure your suggested
payments are in whole amounts of dollars.
Ans. p. 9
2. At a manufacturing plant, plasma screen televisions are made. It is known that each screen has the
chance of being defective of 30%. Size A has a production cost of $1300. If a particular screen is faulty,
this cost is absorbed into the production costs of any non-faulty screens. The daily production is for
8 screens; these are then tested for errors.
If the manufacturer wants to make a 20% profit above the expected production cost of each size A
screen, how much should they each be sold for?
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Use probability distribution models to solve straightforward problems 3
exam
ial
The poisson distribution
notes
Essent
1. A hospital switchboard receives on average 4 incoming calls per minute during the hours of 7 am to 11 pm.
Ans. p. 10
a. Explain why it is likely the number of incoming calls would have a Poisson distribution.
b. Assume the number of calls does have a Poisson distribution. The switchboard becomes
overloaded if more than 6 calls come in during any given minute. When this happens, up to
4 more calls are put on hold and any others are disconnected. Find the probability that if a call
cannot be attended to immediately, it gets put on hold.
2. Plants are delivered to NAILS (a hardware store) in punnets (small containers) by truck. Each
truckload has approximately 250 punnets. The following sampling scheme is used by NAILS to
check on the quality of these punnets.
Scheme 1
A random sample of 20 punnets is chosen from the truckload. If there are fewer than
2 defective punnets in the sample, then the truckload is accepted. If there are at least
2 defective punnets in the sample, then the truckload is rejected.
a. Two per cent of the punnets from one supplier, Grow Well, are known to be defective.
i.
Two truckloads of punnets from Grow Well arrive at NAILS. Calculate the probability that
exactly one truckload of the two is accepted.
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Year 2004
Ans. p. 10
4 Scholarship Statistics and Modelling (Chapter 6)
ii. Under Scheme 1, Grow Well tries to achieve a target acceptance rate of at least 96% of all
truckloads. To the nearest 0.5%, find the greatest percentage defective rate for punnets that
would enable Grow Well to achieve its target acceptance rate.
iii. Suppose that the following two-stage sampling scheme was used.
Scheme 2
A random sample of 20 punnets is chosen from a truckload. If there are fewer than
2 defective punnets in the sample, then the truckload is accepted. If there are at least
2 defective punnets, then a second sample of 20 punnets is chosen. If there are no
defective punnets in the second sample, then the truckload is accepted, but if there are
any defective punnets in the sample, the truckload is rejected.
Under Scheme 2, calculate the proportion of accepted truckloads that would be accepted on
the second sample.
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Use probability distribution models to solve straightforward problems 5
b. Another plant supplier, Quality Plants, has a rate of one defective punnet in 150. Suppose that the
following sampling scheme was used.
Scheme 3
A random sample of n punnets is chosen from a truckload. If there are fewer than
2 defective punnets in the sample, then the truckload is accepted. If there are at least
2 defective punnets in the sample, then the truckload is rejected.
Find the largest value of n that would achieve an acceptance rate of at least 91% (to the nearest
1%) under Scheme 3, of all truckloads from Quality Plants.
exam
ial
The normal distribution
notes
Essent
1. A bottling plant packs bottles of wine in wooden crates. The weights of these crates when empty are
Normally distributed with mean 1 225 g and standard deviation 20 g. Each crate holds 12 bottles. An
empty bottle has a weight that is Normally distributed with mean 155 g and standard deviation 2 g.
Each bottle is filled with an amount of wine that has a Normally distributed weight with mean 763 g
and standard deviation 3 g.
a. Calculate the mean and standard deviation of the weight of a crate of wine.
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Ans. p. 11
6 Scholarship Statistics and Modelling (Chapter 6)
b. What assumption has to be made to calculate the standard deviation in part a? Explain whether
the assumption is reasonable in this situation.
c. Below what weight are the lightest 25% of full crates?
The bottling plant’s quality-control staff weighs every 10th crate to check that the process is working correctly.
d. Draw a box-and-whisker diagram that would be typical of 1 000 consecutive results from the
check, given that the process is working correctly.
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Use probability distribution models to solve straightforward problems 7
exam
ial
Linear combinations of independent random variables
notes
Essent
e. Suppose some bottles of wine are cracked and do not fill with wine. An individual bottle has
probability of π of being cracked, and bottles crack independently. Calculate π correct to 4 sf
given that 4% of the crates fail the above weight test.
1. The variance of a random variable X is defined by the formula
Var(X) = E(X2) – [E(X)]2
Ans. p. 12
If Y = aX + b with a and b constants, prove Var(Y) = a2Var(X).
2. X and Y are independent random variables. It is known that E(X) = 20, E(X2 ) = 500, E(Y ) = 25,
E(Y 2 ) = 750
a. Prove that σX – Y = 15
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Ans. p. 12
8 Scholarship Statistics and Modelling (Chapter 6)
b. If X and Y are Normally distributed find the probability that a randomly chosen value of X will
exceed a randomly chosen value of Y.
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Answers
3.6 The binomial distribution (page 1)
1. a. This is a Binomial distribution with:
1
n = 3, p =
6
b. The formula is used to find the remaining probabilities:
c.
x
1
–1
–2
–3
P (X = x )
125
216
75
216
5
72
1
216
E (X ) = 1 ×
=
5
75
125
1
–3 ×
–2 ×
–1 ×
72
216
216
216
17
216
So after 60 games of this the casino would expect to gain:
17
60 ×
= $4.72
216
d. For a game to be fair, the expected gain for any participant including the casino should be zero.
This can be done in different ways by adjusting the payout amounts for the crowns. One method
is to set one of the payout amounts to be $x.
Let the payout amount for 3 crowns be $x.
5
75
125
1
–x ×
–2 ×
–1 ×
72
216
216
216
x
30
75
125
–
–
–
=
216 216 216 216
20 – x
=
216
For the game to be fair this needs to equal zero. If the payout for 3 crowns is changed to equal
$20 and the rest remains the same, the game is fair.
E (X ) = 1 ×
Another solution would be for the casino to pay $1 for one crown, $3 for 2 crowns and $5 for
3 crowns. This also results in a fair game.
2. Let X = number of defective screens out of 8. X is Binomial with n = 8, p = 0.3 (assuming
independence).
Expected number of defective units = np = 8 × 0.3 = 2.4
Production cost of each saleable unit:
1300 × 8
= $1 857.18
8 – 2.4
Price each should be sold for to get 20% profit:
1 857.18 × 1.2 = $2 228.6
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10 Answers and explanations
3.6 The poisson distribution (page 3)
1. a. Phone calls to the hospital can be assumed to be independent and assuming there is only one
incoming phone line, calls can not occur simultaneously. These are the two conditions for a
Poisson distribution.
b. Require the probability that 7, 8, 9 or 10 calls come in, in any given minute, given that the
switchboard is overloaded.
Let X = the number of calls per minute. X has a Poisson distribution with λ = 4.
P(X ≥ 7)= 0.1107
P(X = 7)= 0.0595
P(X = 8)= 0.0298
P(X = 9)= 0.0132
P(X = 10)= 0.0053
tables or calculator
Hence the probability is:
P ( 7 ≤ X ≤ 10 )
conditional probability formula
P ( X ≤ 10 | X ≥ 7 ) =
P ( X ≥ 7 ) 0.1078
=
0.1107
= 0.9738
2. This question uses the Poisson approximation to the Binomial which is no longer in the Achievement
Standard. The solutions below should be able to be followed by the student.
a. i.
X = number of defective punnets from 20
X is Binomial, n = 20, p = 0.02
P(a truck is accepted) = P(X < 2) = 0.9820 + 20 × 0.9819 × 0.02 = 0.940
P(1 truck accepted out of 2) = 0.94(1 – 0.94) + (1 – 0.94) × 0.94 = 0.113
Alternatively use the Poisson approximation to the Binomial, with λ = np = 20 × 0.02 = 0.4
Probability of an individual truck being accepted:
P(X < 2)= P(X = 0) + P(X = 1)
= 0.6703 + 0.2681
= 0.9384
Probability of exactly one truck being accepted from two:
2 × 0.9384 × 0.0616 = 0.116
ii. Using a Poisson approximation, find λ such that P(X = 0) + P(X = 1) is closest to 0.96. This is
found from tables to be λ = 0.3 hence:
0.3= np
0.3= 20p
p= 0.015
ie greatest rate is 1.5%
iii. This is the conditional probability the truckload is accepted from the second sample given it is
accepted.
Probability of no defective punnets in sample of 20:
0.9820 = 0.6676
Probability of truck being accepted on second check (and rejected on first) is:
0.0616 × 0.6676 = 0.0411
So probability of truck being accepted on the second sample given it is accepted is:
0.0411
= 0.0420
0.9384 + 0.0411
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Answers and explanations 11
b. Using a Poisson approximation, find the value of λ such that P(X = 0) + P(X = 1) is closest to 0.91.
This is found to be λ = 0.5 hence:
So the maximum number of sampled punnets should be 75 to ensure the 0.91 acceptance rate.
λ = np
1
150
n = 75
0.5 = n
3.6 The normal distribution (page 5)
1. a. Let C = weight of crate
Bi = weight of the i th bottle
Wi = weight of wine in i th bottle
12
12
i =1
i =1
T = C + ∑Bi + ∑Wi
E(T)= 1225 + 12 × 155 + 12 × 763 = 12 241 g
SD(T) = Var (T )
= 202 + 12 × 22 + 12 × 32
= 556 = 23.6 g
b. The main assumption is that the weight of each bottle and weight of wine in each bottle are
independent. This is reasonable as the capacity of the bottle does not dictate the amount of wine
poured into it, it is the filling process that controls this.
c. A z-score is required such that the probability of being less than z is 0.25. This happens when
z = –0.674.
x –µ
z=
σ
x – 12 241
–0.674 =
23.6
x = 12 241 – 0.674 – 23.6 = 12 225 g
d. All values are generally accepted to be found within 3.5 standard deviations of the mean, hence
the highest and lowest values can be found:
H = 12 241 + 3.5 × 23.6 = 12 324 g
L = 12 241 – 3.5 × 23.6 = 12 158 g
The lower quartile was found in c. to be 12 225 g
UQ = 12 241 + 0.674 × 23.6 = 12 257 g
As this is a Normally distributed population and the sample number is large, the median can be
taken to be equal to the mean. This gives the following box-and-whisker diagram:
12 100
12 200
12 300
12 400
e. Probability of a particular crate of 12 bottles failing the weight test is 4%, hence 96% pass.
This means P(no cracked bottles) = 0.96
C0(1 – π)12π0 = 0.96
12
1–π=
12
0.96
π = 1 – 0.9966 = 0.0034
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12 Answers and explanations
3.6 Linear combinations of independent random variables (page 7)
1. Var(Y) = E(Y2) – [E(Y)]2
= E(a2X2 + 2abX + b2) – [aE(X) + b]2 since E[aX + b] = aE[X] + b
= a E(X ) + 2abE(X) + b – a [E(X)] – 2abE(X) – b expanding bracket
= a2[E(X2) – [E(X)]2]
= a2Var(X)
2
2
2
2
2
2
2. a. Var(X – Y)= Var(X) + Var(Y)
= E(X2) – [E(X)]2 + E(Y2) – [E(Y)]2
= 500 – 400 + 750 – 625 = 225
σX −Y = 225 = 15
b. X – Y has Normal distribution with mean = 20 – 25 = –5 and standard deviation 15.
(
)
( )
(
)
P (X − Y > 0 ) = P Z >
=P Z >
0 − −5
15
since P(X > Y) = P(X – Y >0)
1
3
= 0.5 − P 0 < Z <
1
= 0.3696
3
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