S2.1 Binomial and Poisson distributions

Document technical information

Format ppt
Size 2.4 MB
First found May 22, 2018

Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

Persons

Organizations

Places

Transcript

A2-Level Maths:
Statistics 2
for Edexcel
S2.1 Binomial and
Poisson distributions
These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page.
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Started presentation.
11 of
of 58
58
© Boardworks Ltd 2006
Binomial distributions
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
22 of
of 58
58
© Boardworks Ltd 2006
Special distributions
Many real-life situations can be modelled using statistical
distributions. Examples of the types of problem that can be
addressed using these distributions include:
In a board game, players needs a six before they can start.
What is the probability that they haven’t started after 5 tries?
What proportion of the adult population have an IQ above
120?
The number of accidents on a stretch of motorway averages
1 every 2 days. How likely is it that there will be no accidents
in a week?
12% of people are left-handed. What is the probability that a
class of 30 people will have more than 6 left-handed people?
3 of 58
© Boardworks Ltd 2006
Binomial distribution
4 of 58
© Boardworks Ltd 2006
Binomial distribution
Introductory example:
A spinner is divided into four equal
sized sections marked 1, 2, 3, 4.
If the spinner is spun 6 times, how likely
is it to land on 1 on four occasions?
One possible sequence would be 1 1 1 1 1′ 1′.
6!
 6C4
The number of possible sequences is
4 !2 !
Most calculators
have an nCr
button
(i.e. the number of ways of arranging 6 items, where 4 are of
one kind and 2 are of a different kind).
Each sequence has probability 0.254 × 0.752.
So the required probability is 6 !  0.25 4  0.752  0.0330.
4 !2 !
5 of 58
© Boardworks Ltd 2006
Binomial distribution
A binomial distribution arises when the following
conditions are met:
an experiment is repeated a fixed number (n) of times
(i.e., there is a fixed number of trials);
the outcomes from the trials are independent of one another;
each trial has two possible outcomes (referred to as
success and failure);
the probability of a success (p) is constant.
If the above conditions are satisfied and X is the random
variable for the number of successes, then X has a
binomial distribution. We write: X ~ B(n , p).
n and p are called
parameters.
6 of 58
© Boardworks Ltd 2006
Binomial distribution
Which of these situations might reasonably be modelled
by a binomial distribution?
1 Joan takes a multiple choice examination
Binomial
consisting of 40 questions. X is the number of
questions answered correctly if she chooses
each answer completely at random.
2
A bag contains 6 blue and 8 green counters.
Not
James randomly picks 5 counters from the bag
binomial
without replacement. X is the number of blue
counters picked out.
Outcomes are not independent
3
A bag contains 6 blue and 8 green counters.
Jan randomly picks 5 counters from the bag,
replacing each counter before picking the next.
X is the number of blue counters picked out.
7 of 58
Binomial
© Boardworks Ltd 2006
Binomial distribution
Which of these situations might reasonably be
modelled by a binomial distribution?
1 Jon throws a dice repeatedly until he obtains
Not
a six. X is the number of throws he needs
binomial
before a six arises.
The number of trials is not fixed
2 Judy counts the number of silver cars
Not
that pass her along a busy stretch of road.
binomial
X is the number of silver cars that pass in
a minute.
The number of trials is not fixed
3 Josh is a mid-wife. He delivers 10 babies.
X is the number of babies that are girls.
8 of 58
Binomial
© Boardworks Ltd 2006
Binomial distribution
9 of 58
© Boardworks Ltd 2006
Binomial distribution
If X ~ B(n , p), then
P( X  x)  nCx p x q n  x
for x  0,1, 2,...n
where q = 1 – p.
Example: X ~ B(12, 0.4).
Find
a) P(X = 3)
b) P(X > 1).
Number
Probability
Probability
of
ofof
x n–x
possible
sequences
successes
failures
12
3
9
P
(
X

3
)

C

0
.
4

0
.
6
 0.142
a)
3
to 3 s.f.
b) P(X > 1) = 1 – P(X = 0) – P(X = 1).
P( X  0)  12C0  0.40  0.612  0.612  0.00218
P( X  1)  12C1  0.41  0.611  0.01741
So P(X > 1) = 0.980 (3 s.f.)
10 of 58
© Boardworks Ltd 2006
Binomial distribution
Example: The probability that a baby is born a boy is 0.51.
A mid-wife delivers 10 babies. Find:
a) the probability that exactly 4 are male;
b) the probability that at least 8 are male.
a) P( X  4)  10C4  0.514  0.496  0.197
b) P( X  8)  P( X  8)  P( X  9)  P( X  10)
  10C8  0.518  0.492    10C9  0.519  0.49   0.5110
 0.04945  0.01144  0.00119
 0.0621
11 of 58
© Boardworks Ltd 2006
Mean and variance of a binomial
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
12 of 58
© Boardworks Ltd 2006
Mean and variance of a binomial
It can be shown that if X ~ B(n, p), then
E[X] = np
E[X] is an unbiased
estimator of the mean.
Var[X] = np(1 – p) = npq.
and
Example:
If X ~ B[16, 0.25], then
E[X] = 16 × 0.25 = 4
and
Var[X] = 16 × 025 × 0.75 = 3
13 of 58
© Boardworks Ltd 2006
Mean and variance of a binomial
Example: If X ~ B(n, p), E(X) = 8 and Var(X) = 4.8,
calculate P(X = 5).
We can use the information provided to form 2 equations:
E[X] = np
so, np = 8
Var[X] = npq
so, npq = 4.8
Substituting the first equation into the second we find 8q = 4.8.
Therefore q = 0.6.
So, p = 0.4 and n = 8 ÷ 0.4 = 20.
Hence, X ~ B(20, 0.4).
20
5
15
So, P(X = 5) = C5  0.4  0.6  0.0746
14 of 58
© Boardworks Ltd 2006
Use of binomial tables
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
15 of 58
© Boardworks Ltd 2006
Use of binomial tables
Tables of probabilities are available for many binomial
distributions. The tables give cumulative probabilities,
that is P(X ≤ x).
x
P(X ≤ x)
0
0.0824
1
0.3294
2
0.6471
P(X ≤ 5) = 0.9962
3
0.8740
P(X = 4) = P(X ≤ 4) – P(X ≤ 3)
= 0.9712 – 0.8740
= 0.0972
4
0.9712
5
0.9962
6
0.9998
P(X > 2) = 1 – P(X ≤ 2) = 1 – 0.6471
= 0.3529
7
1.0000
The table shows an extract for the
cumulative probabilities for a
B(10, 0.3) distribution.
We see that:
16 of 58
© Boardworks Ltd 2006
Use of binomial tables
Example: 1 in 4 people carry a particular gene. If 20 people
are chosen at random, find the probability that:
a) exactly 3 of them carry the gene;
b) at least 6 of them carry the gene.
x
P(X ≤ x)
0
0.0032
The table shows an extract from the
cumulative probabilities for a
B(20, 0.25) distribution. We see that:
1
0.0243
2
0.0913
3
0.2252
a) P(X = 3) = P(X ≤ 3) – P(X ≤ 2)
= 0.2252 – 0.0913
= 0.1339
4
0.4148
5
0.6172
6
0.7858
b) P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.6172
= 0.3828
…
…
17 of 58
© Boardworks Ltd 2006
Use of binomial tables
Examination style question: Jan estimates that
the probability that she has to stay late at work on
any day is 0.2. She plans to keep a record over the
next 16 working days of how frequently she has to
work late. Let X denote the number of such days. x
P(X ≤ x)
0
0.0281
1
0.1407
2
0.3518
3
0.5982
4
0.7983
5
0.9184
6
0.9734
…
…
a) State an assumption needed for a binomial
distribution to be an appropriate model for X.
Assuming that a binomial distribution is
appropriate, find:
b) the probability that she stays late at least
twice;
c) the mean and the standard deviation for the
number of days she will work late.
18 of 58
© Boardworks Ltd 2006
Use of binomial tables
a) The main assumption here would be that the event of
her staying on late at work on any particular day must
be independent of whether she had to work late on
any other day.
Note: The assumption should be
stated in the context of the question.
b) X ~ B(16, 0.2).
P(X ≥ 2) = 1 – P(X ≤ 1)
= 1 – 0.1407 = 0.8593 (from tables)
c) E[X] = np = 16 × 0.2 = 3.2
Var[X] = npq = 16 × 0.2 × 0.8 = 2.56
 s.d = 1.6
19 of 58
© Boardworks Ltd 2006
The Poisson distribution
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
20 of 58
© Boardworks Ltd 2006
Introduction
We are sometimes interested in the number of times an
event occurs in a period of space or time:
1) Sam counts the number of cars travelling past her on a
quiet country road. X represents the number of cars passing
her in 15 minutes.
2) Xiu uses a Geiger counter to record the number of particles
emitted by a radioactive substance. X is the number of
emissions in one minute.
3) Scott counts the number of people leaving a pub. X is the
number of people leaving in a 5 minute interval.
21 of 58
© Boardworks Ltd 2006
Introduction
4) Selina is taking samples of sea water. X is the number of a
particular kind of organism that she finds in a 1 ml sample
of water.
5) Ankur has produced a first draft of a novel. X is the number
of typing mistakes made on a page.
6) Steve records the number of accidents that occur on a
stretch of motorway. X is the number of accidents that occur
in a day.
22 of 58
© Boardworks Ltd 2006
The Poisson distribution
In each of these situations, the random variable X counts the
number of times an event occurs in a given amount of space
or time. X takes the values 0, 1, 2, 3, … .
The Poisson distribution is a model that can sometimes be
used for count data. The distribution is named after the French
mathematician and scientist Siméon Denis Poisson (17811840).
The Poisson distribution has a number of conditions.
23 of 58
© Boardworks Ltd 2006
Conditions for a Poisson distribution
A random variable, X, which counts the number of times an
event occurs in a given unit of space or time will have a
Poisson distribution if:
the events occur independently of each other
and at random;
the events occur at a constant rate (in the sense
that the number of events occurring in a given
interval is directly proportional to the length of that
interval);
the events occur singly (that is, one at a time).
24 of 58
© Boardworks Ltd 2006
The Poisson distribution
The notation used to indicate that a random variable X has a
Poisson distribution is
X ~ Po(λ)
The distribution is fully specified by a single parameter λ,
representing the mean number of events that occur in the given
unit of space or time.
We will now reconsider the seven situations presented
earlier. Decide whether the Poisson distribution might be an
appropriate model in each case.
25 of 58
© Boardworks Ltd 2006
The Poisson distribution
1) The number of cars passing along a quiet
country road in 15 minutes.
2) The number of emissions from a radioactive
substance in one minute.
3) The number of people leaving a pub in a 5
minute interval.
26 of 58
Could be Poisson
Poisson
Not Poisson
© Boardworks Ltd 2006
The Poisson distribution
4)The number of a particular kind of organism
found in a 1 ml sample of seawater.
Could be Poisson
5) Simon has produced a first draft of a novel.
X is the number of typing mistakes made on
a page.
Could be Poisson
6) Steve records the number of accidents that
occur on a stretch of motorway. X is the
number of accidents that occur in a day.
27 of 58
Not Poisson
© Boardworks Ltd 2006
Calculating probabilities
If X ~ Po(λ), then
e  x
P( X = x ) =
x!
for x = 0, 1, 2, 3, …
Suppose X ~ Po(0.85). Find P(X = 3).
e0.85  0.853
P( X = 3 ) =
= 0.0437 (3 s.f.)
3!
28 of 58
© Boardworks Ltd 2006
Calculating probabilities
X ~ Po(0.85). Find P(X > 2).
P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2).
e0.85  0.850
P( X = 0) =
= 0.4274
0!
e0.85  0.851
= 0.3633
P(X = 1) =
1!
e0.85  0.852
= 0.1544
P( X = 2) =
2!
Therefore, P(X > 2) = 1 – 0.9451
= 0.0549
29 of 58
© Boardworks Ltd 2006
Calculating probabilities
On average a call centre receives 1.75 phone calls per
minute.
a) Assuming a Poisson distribution, find the probability that the
number of phone calls received in a randomly chosen
minute is:
(i) exactly 4;
(ii) no more than 2.
b) Find the probability that 6 phone calls are received in a 4
minute period.
30 of 58
© Boardworks Ltd 2006
Calculating probabilities
a) Let X = number of phone calls received in 1 minute.
Then X ~ Po(1.75).
e1.75  1.754
P( X = 4) =
= 0.0679 (3 s.f.)
4!
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
e1.75  1.750
P( X  0 ) 
= 0.1738
0!
e1.75  1.751
P( X  1) 
= 0.3041
1!
e1.75  1.752
P( X  2) 
= 0.2661
2!
Therefore, P(X ≤ 2) = 0.744 (3 s.f.)
31 of 58
© Boardworks Ltd 2006
Calculating probabilities
b) Let Y = number of phone calls received in 4 minutes.
The number of calls in 4 minutes will be on average
1.75 × 4 = 7
So, Y ~ Po(7).
Therefore,
32 of 58
e 7  76
= 0.149 (3 s.f.)
P(Y  6) 
6!
© Boardworks Ltd 2006
Examination-style question
Examination-style question
A gardener has calculated that weeds in his garden occur at a
mean rate of 3.25 per square metre. Assuming that a Poisson
distribution is appropriate:
a) Find the probability that there will be fewer than 4 weeds
in an area of 2 m2.
b) State what needs to be assumed about the distribution
of weeds in order for the Poisson distribution to be fully
justified.
33 of 58
© Boardworks Ltd 2006
Examination-style question
Let X = number of weeds in an area of 2 m2.
a) X = 3.25 × 2 = 6.5, so X ~ Po(6.5).
P(X < 4) = P(X = 0, 1, 2, 3)
e6.5  6.50 e6.5  6.51 e6.5  6.52 e6.5  6.53
=



0!
1!
2!
3!
= 0.00150 + 0.00977 + 0.03176 + 0.06881
= 0.112 (3 s.f.)
b) For a Poisson distribution to be justified, the weeds
would need to occur randomly and at a constant rate.
34 of 58
© Boardworks Ltd 2006
Poisson tables
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
35 of 58
© Boardworks Ltd 2006
Poisson tables
Tables of probabilities exist for many Poisson distributions.
The tables are cumulative, that is they give P(X ≤ x).
λ
0.5
1.0
1.5
2.0
2.5
x=0
0.6065
0.3679
0.2231
0.1353
0.0821
x=1
0.9098
0.7358
0.5578
0.4060
0.2873
x=2
0.9856
0.9197
0.8088
0.6767
0.5438
x=3
0.9982
0.9810
0.9344
0.8571
0.7576
x=4
0.9998
0.9963
0.9814
0.9473
0.8912
x=5
1.0000
0.9994
0.9955
0.9834
0.9580
x=6
1.0000
0.9999
0.9991
0.9955
0.9858
If X ~ Po(1.5), P(X ≤ 4) = 0.9814
36 of 58
© Boardworks Ltd 2006
Poisson tables
λ
0.5
1.0
1.5
2.0
2.5
x=0
0.6065
0.3679
0.2231
0.1353
0.0821
x=1
0.9098
0.7358
0.5578
0.4060
0.2873
x=2
0.9856
0.9197
0.8088
0.6767
0.5438
x=3
0.9982
0.9810
0.9344
0.8571
0.7576
x=4
0.9998
0.9963
0.9814
0.9473
0.8912
x=5
1.0000
0.9994
0.9955
0.9834
0.9580
x=6
1.0000
0.9999
0.9991
0.9955
0.9858
If X ~ Po(1.5), P(X = 2) = P(X ≤ 2) – P(X ≤ 1)
= 0.8088 – 0.5578
= 0.251
37 of 58
© Boardworks Ltd 2006
Poisson tables
λ
0.5
1.0
1.5
2.0
2.5
x=0
0.6065
0.3679
0.2231
0.1353
0.0821
x=1
0.9098
0.7358
0.5578
0.4060
0.2873
x=2
0.9856
0.9197
0.8088
0.6767
0.5438
x=3
0.9982
0.9810
0.9344
0.8571
0.7576
x=4
0.9998
0.9963
0.9814
0.9473
0.8912
x=5
1.0000
0.9994
0.9955
0.9834
0.9580
x=6
1.0000
0.9999
0.9991
0.9955
0.9858
If Y ~ Po(2), P(Y > 1) = P(Y = 2, 3, 4, …)
= 1 – P(Y ≤ 1)
= 1 – 0.4060
= 0.594
38 of 58
© Boardworks Ltd 2006
Examination-style question
Examination-style question
A corner shop has on average 18 customers per hour.
Assume that a Poisson distribution is appropriate.
a) Calculate the probability that
i) more than 10 customers will arrive in a 15 minute interval;
ii) exactly 2 customers will arrive in a 1 minute interval.
b) Find the time interval such that the probability of no
customers arriving during that interval is 0.2.
39 of 58
© Boardworks Ltd 2006
Examination-style question
a) Let X1 be the random variable for the number of customers
arriving in a 15 minute interval.
X1 ~ Po(18 ÷ 4), so X1 ~ Po(4.5).
P(X1 > 10) = 1 – P(X1 ≤ 10)
= 1 – 0.9933 (using tables)
= 0.0067
Let X2 be the random variable for the number of customers
arriving in a 1 minute interval.
X2 ~ Po(18 ÷ 60), so X2 ~ Po(0.3).
P(X2 = 2) = P(X2 ≤ 2) – P(X2 ≤ 1)
= 0.9964 – 0.9631 (from tables)
= 0.0333
40 of 58
© Boardworks Ltd 2006
Examination-style question
b) Let Y be the number of customers arriving in an interval of
length t minutes.
Then Y ~ Po(18t ÷ 60), so Y ~ Po(0.3t).
From the question, P(Y = 0) = 0.2
We can find P(Y = 0) in terms of t:
e0.3t (0.3t )0
= e0.3t
P(Y = 0) =
0!
e0.3t = 0.2
0.3t = ln0.2
ln0.2
t=
= 5.36 minutes
0.3
41 of 58
© Boardworks Ltd 2006
Mean and variance
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
42 of 58
© Boardworks Ltd 2006
Mean and variance
Suppose that X ~ Po(λ).
It can be shown that the mean and variance of X are equal:
E(X) = Var(X) = λ
This result provides us with a useful, informal way to test
whether a variable could be modelled by a Poisson
distribution.
43 of 58
© Boardworks Ltd 2006
Mean and variance
Example: The table below shows the number of goals
scored by each team in matches in the Premiership during
the period from August 21st to September 12th 2005.
r
0
1
2
3
4
5 or more
Frequency, f
21
19
10
3
3
0
Calculate the values of the mean and variance of this
data. Discuss whether these values support the use of
a Poisson distribution as a model for the data.
44 of 58
© Boardworks Ltd 2006
Mean and variance
The mean of the data is:
(0×21)+(1×19)+(2×10)+(3×3)+(4×3) 60
x=
=
=1.071
56
56
Now calculate the variance:

x2 f = (02 ×21)+(12 ×19)+...+(42 ×3) = 134
1
Variance =
n

2
134  60 
2
x f x 
   = 1.245 (4 s.f.)
56  56 
2
It can be seen that the mean and the variance are
approximately equal, suggesting that a Poisson distribution
might be a suitable model for this data.
45 of 58
© Boardworks Ltd 2006
Fitting a Poisson model to data
It is possible to fit a Poisson model to a set of data.
The table below shows the number of goals scored by each
team in matches in the Premiership during the period from
August 21st to September 12th 2005.
r
0
1
2
3
4
5 or more
Frequency, f
21
19
10
3
3
0
Using a Poisson distribution with the same mean as the data,
calculate the theoretical frequencies for 0, 1, 2, 3, 4, or at
least 5 goals in a match.
46 of 58
© Boardworks Ltd 2006
Fitting a Poisson model to data
Let X represent the number of goals scored by a team in a
Premiership match.
The mean of the data was 1.071 goals per match.
We therefore adopt a Po(1.071) distribution to model X.
If X is the random variable for the number of goals scored:
e    r e 1.0711.0710
P(X  0) 

= 0.3427 (4 s.f.)
r!
0!
e 1.0711.0711
P(X =1) =
= 0.3670 (4 s.f.)
1!
e 1.0711.0712
P(X = 2) =
 0.1965 (4 s.f.) etc…
2!
47 of 58
© Boardworks Ltd 2006
Fitting a Poisson model to data
x
P(X = x)
0
0.3427
1
0.3670
2
0.1965
3
0.0702
4
0.0188
5 or more
0.0048
Expected frequencies
P(X ≥ x) is found by
subtracting the sum of the
other probabilities from 1.
48 of 58
© Boardworks Ltd 2006
Fitting a Poisson model to data
x
P(X = x)
Expected frequencies
0
0.3427
19.2
1
0.3670
20.6
2
0.1965
11.0
3
0.0702
3.9
4
0.0188
1.1
5 or more
0.0048
0.3
The expected frequencies
can be found by multiplying
the probabilities by the total
frequency, i.e. 56.
49 of 58
© Boardworks Ltd 2006
Fitting a Poisson model to data
x
f
Expected frequencies
0
21
19.2
1
19
20.6
2
10
11.0
3
3
3.9
4
3
1.1
5 or more
0
0.3
We can see that these expected frequencies are quite
close to the frequencies that were actually observed,
which suggests that the Poisson distribution appears to
be a reasonable model for the data.
50 of 58
© Boardworks Ltd 2006
Approximating a binomial by a Poisson
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
The Poisson distribution
Poisson tables
Mean and variance
Approximating a binomial by a Poisson
51 of 58
© Boardworks Ltd 2006
Approximating a binomial by a Poisson
52 of 58
© Boardworks Ltd 2006
Approximating a binomial by a Poisson
The previous activity showed that there are circumstances
when a Poisson distribution provides a good approximation
to a binomial distribution.
If X ~ B(n, p), then X can reasonably be approximated by a
Poisson distribution with mean np if
Note: It is sometimes
n is large, and
convenient to
approximate a
p is small.
Two frequently used rules of thumb are
n > 50 and np < 5, or
n > 50 and p < 0.1.
53 of 58
binomial with a
Poisson distribution
because it is slightly
easier to calculate
probabilities using a
Poisson distribution.
© Boardworks Ltd 2006
Approximating a binomial by a Poisson
A drug manufacturer has found that 2% of patients taking a
particular drug will experience a particular side-effect.
A hospital consultant prescribes the drug to 150 of her
patients.
Using a suitable approximation calculate the probability that:
a) None of her patients suffer from the side-effects.
b) No more than 5 suffer from the side-effects.
54 of 58
© Boardworks Ltd 2006
Approximating a binomial by a Poisson
Let X represent the number of patients experiencing these
side-effects.
The exact distribution of X is B(150, 0.02).
Since n is large and p is small, X ≈ Po(150 × 0.02)
So, X ≈ Po(3).
e 3  3 0
a) P(X = 0) =
= 0.0498 (3 s.f.)
0!
b) P(X ≤ 5) = 0.9161 (directly from tables).
55 of 58
© Boardworks Ltd 2006
Examination-style question
Examination-style question:
The probability that a directory enquiry service gives out the
correct phone number has been estimated to be 0.975.
a) Sabah requires 10 phone numbers. Find the probability
that the service gives her at least 9 correct numbers.
b) A large organisation requests 140 phone numbers. Find
the probability that more than 135 of them are given out
correctly.
56 of 58
© Boardworks Ltd 2006
Examination-style question
a) Let X be the random variable for the number of correct
phone numbers given to Sabah.
Then X ~ B(10, 0.975).
P(X ≥ 9) = P(X = 9) + P(X = 10).
P( X = 9) = 10C9  0.9759  (1 0.975) = 0.1991
P( X =10) = 10C10  0.97510  (1 0.975)0 = 0.7763
So, P(X ≥ 9) = 0.1991 + 0.7763 = 0.9754
57 of 58
© Boardworks Ltd 2006
Examination-style question
b) The probability of being given the correct phone number
(0.975) is not small.
However, the probability of receiving an incorrect phone
number (0.025) is small.
Therefore we consider Y, the number of incorrect numbers
received.
The exact distribution of Y is B(140, 0.025).
This can be approximated to Po(3.5).
140 × 0.025
The probability of more than 135 correct numbers is
equivalent to the probability of 4 or fewer incorrect numbers.
Using tables: P(Y ≤ 4) = 0.7254
58 of 58
© Boardworks Ltd 2006
×

Report this document