i Educational Publishing � ""'
House Pte
Ltd
-
. Lois Chee
M. Sc., P. G. D. E.
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ChaPter1
Permutations and Combinations ............................................
1
Chanter2
Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Chanter3
Introduction to Statistics . .......................................................
23
Chanter4
Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Cha11er5
Poisson Distribution . ..............................................................
46
Chanter&
Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . .......................................
66
Cha1ter1
Approximation of Distributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
Chanter I
Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
Chanter9
Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
103
Chanter10
Correlation and Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122
Worked Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
151
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Standard discrete distributions
B(n,p)
Binomial
P(X = x)
Mean
e-•-
np
Variance
A.
A.
(;)px(I-p)"-x A.x
Distribution of X
Poisson Po(A.)
x!
np(I-p)
Sampling and testing
Unbiased variance estimate from a single sample: s
2
nI (
=
_
� 2 1 LX
-
-n- ) n I �(x - X)(�)2 =
I
_
2
Regression and correlation
Estimated product moment correlation coefficient: r=
- X")(y -
�(x
Y)
.,j {:2:(x - X")2} {:2:(y - Y)2 }
Estimated regression line of y on x: y
-y
=
b(x - x)' where b =
�(x - X")� -Y) :2:(x - X )2
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@@@@@@@@@@@@@@@@@@@@@@@@@@@@@$@@@@@@@@@@@@@@ @
@
The Normal Distribution Function
If Z has a normal distribution with mean 0 and variance 1 then, for each value of z, the table gives the value of <I>(z), where <I>(z) P(Z:;;;; z). =
For negative values of z use <I>(-z) z
0
2
0.6950 0.7291 0.76ll 0.7910 0.8186
0.6985 0.7324 0.7642 0.7939 0.8212
0.9345 0.9463 0.9564 0.9649 0.9719
0.9357 0.9370 0.9474 0.9484 0.9573 0.9582 0.9656 0.9664 0.9726 0.9732
0.0 O.l 0.2 0.3 0.4
0.5000 0.5398 0.5793 0.6179 0.6554
0.5040 0.5080 0.5438 0.5478 0.5832 0.5871 0.6217 0.6255 0.6591 0.6628
1.0 l.l 1.2 1.3 1.4
0.8413 0.8643 0.8849 0.9032 0.9192
0.8438 0.8665 0.8869 0.9049 0.9207
0.9772 0.9821 0.9861 0.9893 0.9918
0.9778 0.9826 0.9864 0.9896 0.9920
0.5 0.6 0.7 0.8 0.9
1.5 1.6 1.7 1.8 l.9 2.0 2.1 2.2 2.3 2.4
2.5 2.6 2.7 2.8 2.9
0.6915 0.7257 0.7580 0.7881 0.8159
0.9332 0.9452 0.9554 0.9641 0.9713
0.9938 0.9953 0.9965 0.9974 0.9981
3
l
0.9940 0.9955 0.9966 0.9975 0.9982
0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238
4
5
=
1 6
-
<I>(z) 7
.
8
9
0.5160 0.5199 0.5557 0.5596 0.5948 0.5987 0.6331 0.6368 0.6700 0.6736
0.5239 0.5636 0.6026 0.6406 0.6772
0.5279 0.5319 0.5675 0.5714 0.6064 0.6103 0.6443 0.6480 0.6808 0.6844
0.5359 0.5753 0.6141 0.6517 0.6879
0.8554 0.8770 0.8962 0.9131 0.9279
0.8577 0.8790 0.8980 0.9147 0.9292
0.8621 0.8830 0.9015 0.9177 0.9319
0.7054 0.7389 0.7704 0.7995 0.8264
0.7088 0.7123 0.7422 0.7454 0.7734 0.7764 0.8023 0.8051 0.8289 0.8315
0.8461 0.8485 0.8686 0.8708 0.8888 0.8907 0.9066 0.9082 0.9222 0.9236
0.8508 0.8729 0.8925 0.9099 0.9251
0.8531 0.8749 0.8944 0.9ll5 0.9265
0.9783 0.9788 0.9830 0.9834 0.9868 0.987! 0.9898 0.9901 0.9922 0.9925
0.9793 0.9838 0.9875 0.9904 0.9927
0.9798 0.9803 0.9842 0.9846 0.9878 0.9881 0.9906 0.9909 0.9929 0.9931
0.9941 0.9943 0.9956 0.9957 0.9967 0.9968 0.9976 0.9977 0.9982 0.9983
z
0.9382 0.9394 0.9406 0.9495 0.9505 0.9515 0.9591 0.9599 0.9608 0.967! 0.9678 0.9686 0.9738 0.9744 0.9750
0.9945 0.9959 0.9969 0.9977 0.9984
0.9946 0.9960 0.9970 0.9978 0.9984
0.9948 0.9961 0.9971 0.9979 0.9985
l
2
4 4 4 4 4
8 8 8 7 7
12 12 12 ll ll
2 2 2 2 1
5 4 4 3 3
7 6 6 5 4
0.7157 0.7190 0.7224 0.7486 0.7517 0.7549 0.7794 0.7823 0.7852 0.8078 0.8106 0.8133 0.8340 0.8365 0.8389
3 3 3 3 3
0.9418 0.9525 0.9616 0.9693 0.9756
l 1 1 l 1
2 2 2 1 1
0 0 0 0 0
0 0 0 0 0
0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761
0.9441 0.9545 0.9633 0.9706 0.9767
0.9808 0.9812 0.9817 0.9850 0.9854 0.9857 0.9884 0.9887 0.9890 0.99ll 0.9913 0.9916 0.9932 0.9934 0.9936
0.9949 0.9951 0.9962 0.9963 0.9972 0.9973 0.9979 0.9980 0.9985 0.9986
0.9952 0.9964 0.9974 0.9981 0.9986
3
0 0 0 0 0
7 7 6 5 5
l l l l 0
4
16 16 15 15 14
lO lO 9 8 8
14 l3 12 ll 10
4 3 3 2 2
5 4 4 3 2
l l 1 l l
0 0 0 0 0
9 8 7 6 6
2 2 1 1 l
l 0 0 0 0
5
ADD 20 20 19 19 18
6
24 24 23 22 22
7
32 32 31 30 29
36 36 35 34 32
16 14 13 ll 10
19 16 15 l3 ll
21 18 17 14 13
20 19 18 16 15
24 23 21 19 18
6 5 4 4 3
7 6 5 4 4
8 7 6 5 4
2 2 2 1 l
l l 0 0 0
14 12 ll lO 8
3 2 2 2 l
l l l 0 0
9
28 28 27 26 25
17 16 15 14 l3 12 10 9 8 7
8
3 3 2 2 l
l l l 0 0
27 26 24 22 20
10 8 7 6 5 4 3 3 2 2
31 29 27 25 23
ll 9 8 6 5
l l l l 0
Critical values for the normal distribution
If Z has a normal distribution with mean 0 and variance 1 then, for each value ofp, the table gives the value of z such that P(Z:;;; z) =p. p z
0.75
0.674
0.90
l.282
0.95
l.645
0.975
l.960
0.99
2.326
0.995 2.576
0.9975 2.807
0.999
3.090
0.9995 3.291
4 4 3 2 2
l l l l 0
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1.
Basics of Permutations and Combinations
Some students find Permutations and Combinations problems difficult to solve because they have not understood the basic principles. Once they do, Permutations and Combinations actually becomes one of the easiest chapters, because all we need to know is when to use '+', ' ' , 'x' and according to the situation. This chapter will also highlight the common mistakes that we can all learn to avoid. '
-
' +
(a) We need to list all possible cases and add up their numbers when the cases are mutually exclusive. This is called the Addition Principle for counting. E.g. How many code-words of three letters or more can be formed from the word MANSION? Ans: Number of three-letter code-words + Number of four-letter code-words + ... + Number of seven-letter code-words (Note there is n o overlap across the categories.) (b) Instead of adding the valid cases, we can consider the number of cases we do not want and deduct it from the total. This is called the Complementary Principle for counting. E.g. How many ways are there to arrange 10 girls in a row so that the tallest girl does not stand beside the shortest girl? Ans: Number of ways to arrange the girls without restriction - Number of ways to arrange the girls such that the tallest and shortest are together That is, 'what you want = total - what you do not want'. (c) Whenever we consider objects one after the other, we should multiply the number of ways together. This is called the Multiplication Principle for counting. E.g. How many ways are there to seat 10 people around two circular tables, one with 4 seats and the other with 6 seats? Ans: Number of ways to choose 4 people out of 10 x Number of permutations to arrange these 4 people round a circle x Number of ways to choose 6 people out of 6 x Number of permutations to arrange these 6 people round a circle (d) When we arrange objects of which some are identical, we need to divide by a certain factor. There is no specific name for this principle. E.g. How many different whole numbers can be formed from writing the digits 2, 2, 4, 5, 5 and 5 in a row? Ans: Number of ways to arrange six digits in a row (Number of ways to switch the two 2s around x Number of ways to switch the three 5s around) +
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H2 Maths Guide [Statistics]
Anangements If Leners/Peo11e
2.
Some questions will involve restrictions. The most typical restrictions are: (a)
some letters/people must be placed somewhere or together
(b)
some letters/people cannot be placed somewhere or together
Example 1 Find the number of different arrangements of the
10
letters of the word
INCREDIBLE
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in which (i)
all the vowels must be together,
(ii)
none of the vowels are adjacent.
Solution (i)
There are two methods to approach the problem.
Method 1: First, consider the consonants: N,
C, R, D, Band L 6!
Number of ways to arrange the consonants=
Next, consider the vowels. Since they must be together, they will occupy exactly one slot in between the consonants (or at the two ends):
-C-R-D-B-L There are 7 choices The vowels are: I, I, E and E
-N
There are
2��!
'+-j The spaces indicate the possible
-
j pos itions that the vo wels c an
l occupy.
ways of arranging these among themselves in the slot.
Now put it all together.
:. number of arrangements=
6! x 7 x 24,;, = 30 240 1
Method 2:
A
commonly taught way is to think of the
4
vowels as a bundled unit. The
7!
ways to arrange these units
consonants are each a unit.
First, there are
7
units altogether, and
(6 consonants and 1 bundled unit). Next, the number of permutations to :. number of arrangements=
2
arrange vowels within their unit=
7! x 241;1 1
·
·
"'""'''
2��!
"""«' "--' """'«"•"• ' ' ' '"' ""'""
= 30 240 � Notice that the approach is
different but the answer is the �sam e.
:
�
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H2 Maths Guide [Statistics]
&.
Circnlar Per11atatio1s
A distinct difference between seating people on a straight bench and seating them at a circular table is that each person has two 'neighbours' in the latter situation. Also, in a circle, as long as each person retains the same neighbours, they can be rotated round the circle without causing any difference. In general, the number of ways to arrange n distinct objects in a circular manner is (n - 1)!.
At a carnival, 3 boys and 6 girls are to be seated round a carousel with 9 seats. Assuming each child occupies exactly 1 seat, in how many ways can this be done (i) if the 3 boys must sit together, (ii) if 2 particular girls cannot sit together?
(
Solution
(i)
Treat the 3 boys as a bundled unit. Thus there are 7 units altogether (6 girls and 1 bundled unit). Number of permutations for these 7 units = (7 - 1)! = 720 Number of permutations for the 3 boys = 3 ! = 6 Total number of ways = 720 x 6 = 4320
):
(ii) Number of ways without restriction = (9 - 1)! = 40 320 The calcu lation fol lows the Number of ways that the two girls can sit together :. s ame concept as in (i). = (8 - 1)! x 2! = 10 080 Total number of ways = 40 320 - 10 080 = 30 240
...... ... '.. Exantple \
7
路 ---------------------------------------------. -
Same question as above, except that the seats on the carousel are now numbered.
(
Solution
)
Note that there are now 9 distinct seats, thus we multiply the answer in Example 6(i) and 6(ii) by 9. (i) Number of ways = 4320 x 9 = 38 880
(ii) Number of ways = 30 240 x 9 = 272 160 Try it yourself! 8
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Probability
PRACTICE 2
Question 1 [compare with Example 1]
Given that P(A) = 0.3, P(B) = 0.5 and P(A U B) = 0.65. (i) Find P(A n B) and P(A U B). (ii) State, with a reason, whether A and B are independent events. '
Question 2 [compare with Example 2]
Of the cars parked in a particular car park, 70% belong to males and the rest belong to females. 80% of the cars belonging to the males are saloons and 50% of the cars belonging to females are saloons. Additional small side mirrors are fitted to 2% of the non-saloons and 5% of the saloons regardless of whether the saloons are owned by a male or a female. (i) If a car is chosen at random, find the probability that it is (a) a saloon belonging to a female, with additional small side mirrors attached, (b) a non-saloon without additional small side mirrors attached. (ii) A car is chosen at random. Given that it is not a saloon, find the probability that it belongs to a male. Question 3 [compare with Example 3}
In a game show, 10 cards, each labelled with a different number from 1 to 10, are placed in a bag. The game show host first selects 3 cards at random, without replacement, from the bag. A contestant next chooses 4 different numbers from 1 to 10 at random. The numbers on the cards selected by the game show host are then revealed to the contestant, who will match his chosen numbers with the numbers on the cards. (i) Find the probability that the contestant has exactly 3 matching numbers. (ii) Given that the contestant has at least 1 matching number, find the probability of having exactly 3 matching numbers. Question 4
Given that P(A) = 0.5 and P(Bďż˝ = 0.3, determine whether A and B can be mutually exclusive. Question 5
Jane has 20 different pairs of shoes. Assuming that the number of left and right shoes need not be in equal numbers, find the probability that there is no matching pair of shoes if she chooses 12 shoes at random.
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H2 Maths Guide [Statistics]
3 .2 For any random variable X with E(X) = µ, the variance of X, Var(X), is given by: Var(X) = E(X - µ)2
Alternatively, Var(X) = E(X2) - µ2 . 3 .3 Var(X) is also denoted by the symbol d2. 3.4 Note that Var(X) is a non-negative value. 3.5
u=
VVar(X)
= standard deviation
of X
3.6 For any random variable X, (a) Var(a) = 0 (b) Var(aX) = a2Var(X) (c) Var(aX + b) = a2Var(X), where a is a constant. 3. 7 For any random variables XI> X2, . . . , Xm
3.8 For independent random variables XI> X2, . . . , Xm
I Var(X1 + X2 + . . . + Xn)
=
Var(X1) + Var(X2) + . . . + Var(Xn)
3.9 For random variables X and Y and constants a and b, (a) E(aX + bY) = aE(X) + bE(Y) (b) E(aX - bY) = aE(X) - bE(Y) 3.10 If X and Y are independent, (a) Var(aX + bY) = a2Var(X) + b2Var(Y) (b) Var(aX - bY) = a2Var(X) + b2Var(Y) i
26
�
- -
_L
_
__ _ _
N ote that this is a + sign.
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1.
Bino1Dial DisUibution
1 . 1 If a discrete random variable X is the number of 'successes' out of n trials, it follows a binomial distribution given by P(X = r) = ( ; ) pr(I - py - r, for r = O, 1, 2, 3, . . . , n
where n is the number of trials and p is the probability of 'success'. 1 .2 Notation: X - B(n, p) 1 .3 The binomial variable X can take the values 0, 1, 2, 3, . . . , n. 1 .4
Expected value, µ =
1 .5
Variance,
E(X) = np
a2 = Var(X) = npq, where q = 1 - p
If X - B(n, p), using the graphic calculator, to find P(X = r): Step 1 : Press [2ND][VA RS]. Step 2: " Select A:binompdf( and key the values in the format n, p, r. •
•
2.
to find P(X :::::;; r): Step 1 : Press [2ND][VARS]. Step 2: Select B:binomcdf( and key the values in the format n, p, r.
BiDOIDial Distribution ISSUIDPtions
2. 1 The experiment has n repeated and independent trials. 2.2 There are only two possible outcomes, 'success' and 'failure', for each trial. 2.3 The probability of success, p, remains constant over all trials. © Educational Publ ish i ng House Pte Ltd
29
H2 Maths Guide [Statistics]
&.
Setting a Binomial Di•ibuUon over inother Binomial Di•ibUUon
Example
6
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Past records show that 3 out of 8 customers at a supermarket bring their own grocery bags. If a sample of 12 customers are selected at random from the customers in the supermarket, determine the probability that (i) more than 2 customers bring their own grocery bags, (ii) more than 2 customers bring their own grocery bags given that at most 8 of the customers bring their own grocery bags, (iii) at least 10 of the samples have at most 2 customers bringing their own grocery bags, given that 20 such samples are taken. Solution
(i)
Let Xbe the number of customers who bring their own grocery bags in a sample of 12 customers. The success outcome for X is X - B ( 12, ! ) defined as customers bringing their own grocery bag.
P(X > 2) = P(X ;;;i: 3) = 1 - P(X :s:: 2) = 0·886 !
(ii)
• 1 - binomcdf(1 2, 3/8, 2
Method 1:
P(X > 2 I x :::::; 8) = P(X >P(X2 n...x8)..: 8) P(3 ..: x ..: 8) P(X ..: 8)
=
Method 2 :
P(X ..: 8) - P(X ..: 2) P(X ..: 8)
0.885
II(
. qJ .. .
• (binomcdf(l 2, 3/8, 8) - binomcdf(12, 3/8, 2))/binomcdf(1 2, 3/8, 8)
P(X > 2 I x � 8) = P(X >P(X2 n...x8)..: 8) =
P(3 ..: x ..: 8) P(X ..: 8)
0.885
l.
• sum(binompdf( 1 2, 3/8, {3, 4, 5, 6, 7, 8}))/binomcdf(12, 3/8, 8) 38
_.
...
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Poisson Distribution
Example
1
A random variable X has a Poisson distribution with P(X = 3) = 0.1 . (i) If the mean µ is such that µ > 5, state the mean µ and the standard deviation a, giving your answer correct to 3 significant figures. (ii) Using your answer in (i), find P(if < X < µ + a). Solution
X - Po(µ) (i)
P(X = 3) = 0. 1
From the graphic calculator, we obtain the graph below. y
-+-
Always remember to sketch the : graph as part of the essential working even if the question does ' not explicitly ask for it.
Since µ > 5, µ = 5.77 (reject µ = 1 .302). For Poisson distributions, the mean µ is equal to the variance cr2. Hence, ,; = 5.768 <7 = 2.40
(f/f There are two methods of using the graphic calculator to sketch the above graph and obtain a point of intersection:
Method 1:
Step 1 : Press [Y=]. e-xxJ Step 2: At Y 1 and Y2, enter 31 and 0. 1 respectively. Step 3: Press [2ND][TRA CE] and select 5:intersect to sketch the graph. Step 4: To obtain a point of intersection, press [ENTER] three times.
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H2 Maths Guide [Statistics J
(1 z1�lutio�
1J
Let C be the number of requests for the hire of a car in a day. C - Po(2.8) (i)
P(a request for the hire of 3:..��,�� t�, ��,i!�}��.2�.�---�1L.�. = P(C > 4) The garage only has 4 cars. If there are more than 4 requests received in a day, the request for the hire of a car for that day = 1 - P( C :s:;; 4) has to be rejected. = 0. 152
-.,.�,..,_-....�·"-·- -··· ·
L i
""""!!ili!!il.� .� _¥,- 0� - � -� �� - ��1111!llmlil 11Wl! ffl'l l �!llli�lilll i'iil' Olll lll!'l �!!11!\l ll!!1 B'!l!lll l11 .,
1 - poissoncdf(2.8, 4)
(ii) Let n be the least number of cars in the garage such that P(a request for the hire of a car for that day has to be rejected) < 0.05 P(C > n) < 0.05 1 - P(C :s:;; n) < 0.05 P(C :s:;; n) > 0.95 From the graphic calculator, n
P(C :s;; n)
4
0.84768
5
0.93489
6
0.9755 9
Similar to Example 2, first use a graphic calculator to generate a list of values of P(X ..;; r) for different values of r. Next, determine the smallest value of r such that P(X ..;; r) > 0.95.
Hence the least value of n = 6. 6-4=2 Therefore, the garage owner needs to buy at least 2 more cars.
� Try it yourself! Go to Question 3 . 3. eondlUons tor Assuming a Poisson Model
3.1 The average number of occurrences is constant for each interval, i.e. the mean number of events occurring in any interval (of time or space), is proportional to the size of the intervals. 3.2 The probability of more than one occurrence in a given short interval is very small. 3.3 The number of occurrences in disjoint intervals is independent of each other. 50
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H2 Maths Guide (Statistics)
5. Problems Involving Poisson and Binomial Distrib•on
Very often, a problem may involve both a Poisson and binomial distribution. For such cases, always define the Poisson and binomial random variables using different letters.
Example
6
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In a publishing company, typographical errors are found on the pages of the draft of a book. On average, 2.5 typographical errors are found per 100 pages. If a page contains one or more errors, it will be sent for correction. (i) A staff is required to estimate the percentage of pages that have to be sent for correction. Consider the following statement: 'Since 100 pages contain 2.5 typographical errors each, approximately 2.5% of the pages will have to be sent for correction.' State, giving a reason, whether the statement is valid. (ii) Find the probability that (a) a page will be sent for correction, (b) 2 out of 3 pages will be sent for correction, (c) there are less than the average number of typographical errors on 80 pages.
Solution (i)
The statement is not valid because each page may contain more than one typographical error.
(ii)
(a) Let X be the number of typographical errors on a page. X - Po(0.025) P(a page will be sent for correction) = P(X � 1)
: � . �2�c:;_o_) ...
liF=�=�;.25:�)-) ;J
__
W iti e .IWIU.. l .li iiu
liii) £$ $Z !#Z
(b) Let Y be the number of pages, out of 3 pages, that will be sent for correction. t"- -� Note that the random variable is ..
-... -...,::,.--..,.;..�--·"'-·"'"'"" ·''' '"�
-
Y - B(3, 0.02469)
. _ ,__
t
. . . . . . . . . ....., · · · ·- - · --�-- ·----·- ' · - ·
.......
..
l
...... .·�--�-·,,,,...,;;,.,.,,,....,
\ now different.
. . ... . . .·
Y follows a binomial distribution with n = 3. The probability, p, that a page will be sent for correction can be obtained from (ii).
56
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H2 Maths Guide (Statistics]
10. PllSSll Distribldll PrablllDS IDVllVllU Permlll8tions 8111 C11Dbi1ad11s
ExaD1.ple
11
... ___________________________ a ' ,_..________....__
On average, a randomly chosen sales manager closes 1 business deal per fortnight, where each business deal is independent of other similar deals. Assuming that a month consists of 2 fortnights, find the probability that (i) 3 such sales managers close at least 4 business deals in a period of 1 month, (ii) of 3 such sales managers, 1 closes exactly 4 business deals and the other 2 do not close any business deals in a fortnight.
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[• Solu�!�:x ) (i)
Let X be the number of business deals closed by a sales manager in a period of 1 month. X - Po(2)
P(X1 + X2 + X3 ;;J!: 4)
--
1 P(X1 + X2 + X3 < 4) 1 P(X1 + X2 + X3 � 3) = 0.849·�------; =
=
1
-
poissoncdf(6,
3)
(ii) Let Y be the number of business deals closed by a sales manager in a period of 1 fortnight. Y - Po(l) Required probability = P(Y1 = 4)P(Y2 = O)P(Y3 = 0) = 0.00622
. · --·��.·-"�··-··· · · · ·
x
. . ..
. .. . .
3 � Remember to multiply by 3 since there are 3 ways of arranging: i (4, 0, 0) , (0, 4, 0), (0, 0, 4). i It is in fact the same as finding the number of ways to arrange a '4' and two 'O's in a row.
I
J ·
I/it
poissonpdf( l , 4)(poissonpdf( l ,
0))2
x
3
� Try it yourself! Go to Question 13. 62
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1.
Normal Dlstribntlon
1 . 1 If X has a normal distribution with mean, E(X) = µ and variance, Var(X) = 02, X - N(µ, 02) 1 .2 X is a continuous distribution. 1 .3 The probabiliy density function of a normal distribution can be represented by a bell-shaped curve, that is symmetrical about the mean, µ.
µ 1 .4 The probability for a normal distribution is equal to the area under the normal distribution curve. P(a .s;; X .s;; b)
a
a
a
b
P(X = a) = 0 since the corresponding area under the curve is zero. Hence for normal random variables, we have P(X .s;;; a) = P(X < a) + P(X = a) = P(X < a) Similarly, P(X ;:;:,: a) = P(X > a) + P(X = a) = P(X > a) Total area under normal curve = Total probability = 1
66
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H2 Maths Guide [Statistics]
( Solution .
I I J22 t
Given: X - N(µ, 32)
(i)
Since the normal curve is symmetrical about µ, and P(X < 3) = P(X > k), µ is the midpoint of 3 and k. 3 +k => µ = -- (shown) 2
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The two shaded areas are equal
'
(ii)
3
P(X > 4) = 0.8 => P(X � 4) = 0.2 =>
=>
µ
k '
P{z � 4 ; µ ) = 0.2+-l Standardise �ince we do not kno� �e value of µ .
µ = 6.52486
..________
+k Substitute the value ofµ into µ = 3� : 3 k 6.52486 = ; :. k = 10.0
(iii) P(X �
�)
=
=
P(X � 5.02486) 0.691 \
Take llo�e
l..
l/Jf;.. invNorm(0.2)
'l4fT
l/j normalcdf(S.02486, 6.52486, ..
E99,
.
3)
(a) In order to use the invNorm function of the graphic calculator, we need the probability to be of the form P(X < a) or P(X � a). Hence (i) change P(X > a) to 1 - P(X � a), (ii) change P(X � a) to 1 - P(X < a) . (b) To use the normalcdf or invNorm functions of the graphic calculator, note that we must enter the value of the standard deviation u, NOT the variance a2 into the graphic calculator.
'3Ji Try it yourself! Go to Question 1 . 68
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H2 Maths Guide (Statistics]
Since n = 280 > 50, np = 56 > 5, n( l - p) = 224 > 5, Y - N(56, 44.8) approximately.
t._
___,.
______
npq
=
56 (1 - 0.2) 44.8 x
=
E(X + Y) = 10 + 56 = 66 Var(X + Y) = 9.5 + 44.8 = 54.3
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=> X +
Y - N(66, 54.3) approximately
P(X + Y � 70) � P(X + Y < 70.5) = 0.729
i
Sum up the approximated normal distributions in order to estimate the : required probability.
..,_,
_____..,..
(ii)
normalcdf(-E99, 70.5, 66, v54.3)
E(4X) = 4 x 1 0 = 40 Var(4X) = 42 x 9.5 = 1 52 => 4X - N(40, 1 52) approximately E(Y - 4X) = 56 - 40 = 1 6 Var(Y - 4X) = 44.8 + 1 52 = 1 96.8 => Y - 4X - N(1 6, 1 96.8) approximately P(Y > 4X) = P(Y - 4X > 0) � P(Y - 4X > 0.5) = 0.865
-----
normalcdf(0.5, E99, 16, vl96.8)
Try it yourself! Go to Question 3 . 4.
Normal APProximauon to the Poisson DistribuUon
4. 1 Suppose we are given the random variable X, which follows a Poisson distribution, X - Po(µ). Ifµ > 1 0, then X - N(µ, µ) approximately. 4.2 Follow the steps below in answering such questions. Step 1 : Define the Poisson random variable. Step 2: Indicate X - Po(µ). Step 3: Justify the use of normal approximation. Step 4: Write down the approximated normal distribution. Step 5 : Perform continuity correction and proceed to find the required probability.
86
-----
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H2 Maths Guide [Statistics]
Example
6
The marks obtained by a random sample of eight students in two English tests, x and y, are shown in the following table.
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Student Test 1
mark, x
Test 2
mark, y
(i)
(ii)
1
2
3
4
5
6
7
8
16 18
19 20
14 10
17 17
9 8
19 23
21 20
15 14
Find, in any form, the equation of the regression line of (a) y on x, (b) x on y. Jane scored 1 3 in Test 2 but missed Test 1 . Use the appropriate line to estimate what her mark (to the nearest whole number) for Test 1 should be. Is this estimated value reliable? Why?
Solution (i)
(ii)
From the graphic calculator, (a)
regression line of y on x is y = -4.33 + l .27x,
(b)
regression line of x on y is x = 5.66 + 0.652y.
We use the regression line of x on y since we are predicting the value of x from
y. When y = 13, x = 5.6596 + 0.65 1 72(1 3)
14. 1 32 "' 14 =
Hence, Jane's estimated mark for Test 1 is 14. From the graphic calculator, rxy = 0.909 which is close to + 1. This indicates that there is a strong linear correlation between x and y, thus the regression line is a good fit for the data. Moreover, y = 1 3 is within the data range 8 :::::; y :::::; 23. Thus the estimate is reliable.
Try it yourself! Go to Question
1 30
5.
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H2 Maths Guide [Statistics)
(iii) The value of y = a = 0.785 is obtained when x = 0. However x = 0 falls outside the input data range 40 � x � 56, and the linear relationship may not hold outside the data range. Hence the interpretation of the value of a may not be valid
,. ,. .- .,
...,_;/ is in terms of thousands) .. . � - · - ··· -
(iv) When x = 70, y = 0.78504 + 0.037942(70) = 3 .44
· - .
?J>ii<?AWil*kk·��e•.<!U. ·M,�'''"J:f:'.kJi!W,,.\W!W'i.$.• .
Hence 3440 bottles will be sold in a particular month when the amount of advertising time on radio in that month is set to be 70 minutes.
Try it yourself! Go to Question 1 1 .
9.
Method 11 Least sauares
9 . 1 The method of least squares is used to find the line of best fit or the regression line. 9 .2 To find the equation of the regression line of y on x: y = a + bx. Step 1 : Plot the given data points (x" y1), (x2 , y2), . , (xn , Yn) on a scatter diagram. Here, x is treated as the independent variable and y as the dependent variable. Step 2: Draw a straight line, as close as possible to the data points. Step 3 : Let e" e2 , , en be the vertical difference from each data point to the line. Step 4: Find the values of a and b such that the value of "i:.e/ is as small as possible. Note that for each i = 1 , 2, . . . , n, ei = Yi - a - bxi. • •
• • •
y
1 40
(x.,y.)
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Correlation and Regression
Question 10 [compare with Example 1 1]
A random sample of five pairs of values of x and y is used to obtain the following equations of the regression lines of y on x and x on y respectively: y = -0.81 506 + 1 .40987x, (i)
x = 1 . 1 653 1 + 0.69235y.
Show that x = 25.2 and y = 34. 7, correct to one decimal place.
Five pairs of data are given in the table.
(ii)
x
26
29
23
27
11
y
37
40
33
39
13
Using (i), find the sixth pair of values of x and y, giving your answer correct to one decimal place.
Question 1 1 [compare with Example 12}
The following data shows the yield y (in grams) of a chemical reaction run at various temperatures () (in °C).
(i) (ii) (iii) (iv)
(}
90
1 00
1 10
120
1 30
140
1 50
y
3.9
5.0
4.5
3.8
4.5
3.1
1 .4
Find the equation of the regression line of y on (), in the form y = a + b(). Interpret the values of a and b in the context of the question. Explain why the interpretation of the value of a is not valid. Calculate the amount of yield that would be produced when the temperature at which the chemical reaction is carried out is 125°C.
Question 12 [compare with Example 13]
A random sample of six pairs of values of x and y is recorded as follows. x
26
29
23
27
11
35
y
37
40
33
39
13
46
Let X be the value obtained by substituting a sample value of y into the equation of the regression line of x on y. (i) Evaluate X for each of the six values of y and find the value of �(x X)2 • (ii) For each of the five sample values of x, X' is given by X' = A + By, where A and B are any constant. Deduce an inequality involving �(x X')2. -
-
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Worke d S olutions
Question 1
(i)
Given P(A) = 0.3, P(B) = 0.5, P(A
U
B) = 0.65
By the Inclusion-Exclusion Principle, P(A U B) = P(A) + P(B) - P(A n B) => 0.65 = 0.3 + 0.5 P(A n B) => P(A n B) = 0. 1 5 -
(ii)
S .------.
B
A
From the Venn diagram, P(A ' U Bďż˝ = 1 P(A n B) = 1 - 0. 1 5 = 0.85 -
Since P(A)
x P(B) = 0.3 x 0.5 = 0. 1 5 = P(A n B), events A and B are independent.
Question 2
Let M denote the event that the car belongs to a male. Let S denote the event that the car is a saloon.
(i)
(a)
P(a saloon belonging to a female, with additional small side mirrors)
= 0.3 x 0.5 x 0.05 = 0.0075
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Binomial Distribution
Method 2: (Graphic calculator method)
P(Y � 1) � 0.975 => 1 - P(Y = 0) � 0.975 => P(Y = 0) :s;; 0.025 From the graphic calculator, when n = 3, P(Y = 0) = 0.0353 when n = 4, P(Y = 0) = 0.0 1 1 6 when n = 5, P(Y = 0) = 0.00380 Therefore, the least value of n is 4. (iii)
Admitted to the ICU Not admitted to the ICU 0.4 Admitted to the ICU 0. 1 Not requiring blood transfusion � � Not admitted to the ICU
3
Requiring blood transfusion
g
06
( ! )co.6) + ( ! )co .1)
P(a randomly chosen patient is admitted to the ICU) =
= 0.2875
Let W be the number of patients who are admitted to the ICU out of 200 patients. W - B(200, 0.2875) P(at most 76 patients are admitted to the ICU) = P(W � = 0.998
76)
Question 7
(i)
Although the 1 50 beans are drawn without replacement, the large number of beans in the huge pot means that the proportion of red beans and green beans is approximately constant.
P(drawing a green bean)
32 ... 68+32
= 0.32 approximately Therefore, X can be very well approximated by a binomial distribution. X - B(1 50, 0.32) approximately
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Worlt e d S o lutions
Question 1
X - Po(u) (i)
P(X = 2) = 0. 1 5
e-µµ2
� 21 = 0. 1 5
From the graphic calculator, we obtain the graph below. Since µ > 3, µ = 3.953 Hence, µ = 3.95 correct to 3 significant figures. rl- = 3 .953 (J = 1 .988 Hence, u = 1 .99 correct to 3 significant figures.
y
(ii)
P(X � µ u) = P(X � 3 .953 1 .988) = P(X � 1 .965) = P(X � 2) = 1 - P(X < 2) = 1 - P(X � 1 ) = 0.905 -
-
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