Section 14.1 Random Variables
CHAPTER 14
DISCRETE PROBABILITY DISTRIBUTIONS
(page 291)
1.(a)The number X of industrial accidents per year is discrete.
The possible values are 0, 1, 2, 3, . . . .
(b)The length of time Y to type 1 000 words is continuous.
The range is (0, ).
(c)The number Z of flaws in 1 m2 of cloth is discrete.
The possible values are 0, 1, 2, 3, . . . .
(d)The amount T of oxygen produced in a chemical reaction is continuous.
(The amount is any volume within certain range.)
2.(a)P is continuous.
(b)M is discrete.
(c)U is discrete.
(d)V is continuous.
3.(a)P is discrete. Its range is {0, 1, 2, 3, . . . }.
(b)T is continuous. Its range is [0, ).
4.(a)N is discrete.
Depending on the size and complexity of the model, N can assume values 1, 2, 3, …
(b)W is continuous.
Depending on the size and complexity of the model, W can assume values in the interval
0 < W.
5.(a)The range of P = {0, 0.02, 0.04, 0.06, . . . , 0.98, 1}
(b)The range of G = {1, 2, 3, 4, . . . }
6.(a)X can assume values 0, 1 and 2.
(b)Let the sample points be denoted by (NN), (NY), (YN) and (YY), where N represents a “no” response and Y a “yes” response. The first letter represents the response of one of the two pedestrians and the second letter represents the response of the other pedestrian.
The following table gives the probabilities associated with the various values of X.
Sample point / Value of X / Probability(NN) / 0 /
(NY), (YN) / 1 /
(YY) / 2 /
7.(a)T can assume values 1.5, 2.5 and 3.
(b) / P(T > 2) = P(T = 2.5 or 3)= P(T = 2.5) + P(T = 3) / Addition rule for mutually exclusive events
=
=
=
8.(a)The sample space is
{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH,
{THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}.
(b)(i)
X(e) =
(ii) / x / 0 / 1 / 2 / 3 / 4P(X = x) / / / / /
9.(a)The range of D is {5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5}.
(b)Let (x, y) denote the outcome with the green die scoring x and the red die scoring y.
(i)The outcomes for the event D = 3 are: (4, 1), (5, 2), (6, 3).
(ii)The outcomes for the event D = 2 are: (1, 3), (2, 4), (3, 5), (4, 6).
(c)(i)P(D = 3) =
(ii)P(D = 2) =
10.(a)The possible outcomes are:
(red, red), (red, yellow), (yellow, red), (yellow, yellow).
(b)
(c)The mappings of Y and R are not identical.
For example,Y ( (red, red) )= 0
butR ( (red, red) )= 2.
11.(a)Let D denote a defective item and N denote a non-defective item.
The sample space = {DDD, DDN, DND, NDD, DNN, NDN, NND, NNN}.
(b)X(DDD)= 3
X(DDN)= X(DND) = X(NDD) = 2
X(DNN)= X(NDN) = X(NND) = 1
X(NNN)= 0
(c)P(X = 2) =
12.(a)Range of H = {0, 1, 2, 3, 4, 5}
(b)P(H = 1) = = 0.411 4
(c)Range of K = {0, 1, 2, 3, 4}
(d)P(K > 1)= 1 – P(K = 0) – P(K = 1)
= 1 –
= 1 – 0.658 84 – 0.299 47
= 0.041 7 (4 d.p.)
13.(a)The range of Z = {3, 4, 5, . . . , 17, 18}
The range of N = {0, 1, 2, 3}
(b)(i)The sample points for Z 4 are:
(1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1).
(ii)The sample points for N = 2 are:
(1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 1, 4), (1, 4, 1),
(4, 1, 1), (1, 1, 5), (1, 5, 1), (5, 1, 1), (1, 1, 6), (1, 6, 1), (6, 1, 1).
(iii)The sample points for Z 4 and N = 2 are:
(1, 1, 2), (1, 2, 1), (2, 1, 1).
(c)P(Z 4 and N = 2) =
14.(a)The sample points are:
ABC, ACB, BAC, BCA, CAB, CBA
where ABC denotes the case in which circular A is delivered to the accounting department, B to the production department and C to the marketing department and so on.
(b)(i)The range of X = {0, 1, 3}
(ii)The outcomes which correspond to X = 1 are: ACB, CBA, BAC.
(iii)The outcomes which correspond to X = 0 are: BCA, CAB.
There are 5 outcomes corresponding to X 1.
P(X 1) =
15.(a)The number of possible outcomes == 15
(b)
X(e) =
x / 0 / 1 / 2P(X = x) / / /
(c)
Y(e) =
y / 3 / 3.5 / 4 / 4.5 / 5 / 5.5 / 6P(Y = y) / / / / / / /
(d)The outcomes corresponding to X = 2 and Y = 4 are {Adam, David} and {Brian, Fred}.
P(X = 2 and Y = 4) =
1
Section 14.2 Probability Distributions and Probability Functions
EXERCISE 14.2 / Section 14.2 Probability distributions and probability functions(page 297)
1.Not valid, because the sum of the probabilities = 1.1 ( 1).
2.Valid, because f (x) > 0 and = 1.
3.Not valid, because f (5) < 0.
4.Valid, because f (x) > 0 and = 1.
5.(a)Since = 1,
0.15 + k + 0.2 + 0.25 + 0.3= 1
k= 0.1
(b)
(c)P(X 6)= P(X = 2) + P(X = 4) + P(X = 6)
= 0.15 + 0.1 + 0.2
= 0.45
6.(a)= 1
k + 2k + 3k + 4k= 1
k= 0.1
(b)(i)P(Y 3) = f (1) + f (2) + f (3) = 0.1 + 0.2 + 0.3 = 0.6
(ii)P(1.5 Y 3) = f (2) + f (3) = 0.2 + 0.3 = 0.5
(c)
7.(a)Since all the bars are of equal height and the sum must be equal to 1, each bar must represent a probability of . The probability distribution of X is given below.
x / 10 / 12 / 14 / 16 / 18f (x) / / / / /
(b)Let f (x) be the probability function.
Then the probability distribution is given by
(c)P(12 X < 16)= P(X = 12) + P(X = 14)
=
=
8.(a)Since the total area of the rectangles = 1,
k + 2k + 2k + k= 1
k=
The probability distribution of Y is given below.
y / 0 / 1 / 2 / 3f (y) / / / /
(b)P(Y > 1.5)= P(Y = 2) + P(Y = 3)
=
=
9.(a)Since = 1,
k=
(b)f (t) = ,fort = 0, 1, 2
The probability distribution of T is given below.
t / 0 / 1 / 2f (t) / / /
10.(a)Since= 1,
= 1
= 1
= 1k= 4
(b)P(W > 4)= 1 – P(W 4)
= 1 –
=
11.(a)The sample space is
S = {TT, TH, HT, HH},
where T = tail, H = head and S is equiprobable since the coins are honest.
(i)The probability distribution of X is shown below.
x / 0 / 1 / 2f (x) / / /
(ii)The following bar chart shows the probability distribution of X.
(b)(i)The probability of a head (H) occurring is now while that of a tail (T) occurring is
.
The sample space S = {TT, TH, HT, HH} is no longer equiprobable.
P(TT) =
P(TH) =
P(HT) =
P(HH) =
The probability distribution of X is now:
x / 0 / 1 / 2f (x) / / /
(ii)The following bar chart shows the distribution.
12.(a)The possible outcomes and the associated sums T are:
(7, 7); 14, (7, 8); 15, (7, 9); 16, (8, 7); 15, (8, 8); 16,
(8, 9); 17, (9, 7); 16, (9, 8); 17, (9, 9); 18.
The probability distribution of T is:
t / 14 / 15 / 16 / 17 / 18f (t) / / / / /
(b)The possible outcomes and the associated sums T are:
(7, 8); 15, (7, 9); 16, (8, 7); 15, (8, 9); 17, (9, 7); 16, (9, 8); 17.
The probability distribution of T is:
t / 15 / 16 / 17f (t) / / /
13.(a)The probability distribution of G is:
g / 0 / 1 / 2 / 3f (g) / (0.6)3x
= 0.216 / 3(0.4) (0.6)2
= 0.432xxxx / 3(0.4)2 (0.6)
= 0.288xxxx / (0.4)3
= 0.064
(b)P(G 1) = P(G = 0) + P(G = 1) = 0.216 + 0.432 = 0.648
14.(a)P(X = 0) =
P(X = 1) =
P(X = 2) =
P(X = 3) =
The probability distribution of X is:
x / 0 / 1 / 2 / 3f (x) / / / /
(b)P(X 1) = 1 P(X = 0) =
15.(a)P(X = 0) =
P(X = 1) =
P(X = 2) =
P(X = 3) =
P(X = 4) = 0
The probability distribution of X is tabulated below.
x / 0 / 1 / 2 / 3 / 4fX(x) / / / / / 0
(b)SinceX + Y = 4,Y = y is the same as X = 4 – y.
P(Y = y) = P(X = 4 – y),fory = 0, 1, 2, 3, 4
The probability distribution of Y is thus shown below.
y / 0 / 1 / 2 / 3 / 4fY(y) / 0 / / / /
1
Section 14.3 Expectation
16.(a)The required number = 9 10 10 10 = 9 000
(b)P(X = 0) =
P(X = 1) =
P(X = 2) =
P(X = 3) =
P(X = 4) =
The probability distribution of X is:
x / 0 / 1 / 2 / 3 / 4f (x) / / / / /
EXERCISE 14.3 / Section 14.3 Expectation
(page 306)
1.(a)E(X)=
= 0 0.2 + 1 0.4 + 2 0.3 + 3 0.1
= 1.3
(b)(i)The probability distribution of Y is shown below.
y = x + 2 / 2 / 3 / 4 / 5fY(y) / 0.2 / 0.4 / 0.3 / 0.1
(ii)From the probability distribution in (i),
E(Y)=
= 2 0.2 + 3 0.4 + 4 0.3 + 5 0.1
= 3.3
(iii)SinceY= X + 2,
E(Y)= E(X) + 2
= 1.3 + 2,from the result in (a)
= 3.3
2.(a)E(W)=f (w)
= 1 0.1 + 3 0.2 + 5 0.4 + 7 0.2 + 9 0.1
= 5
E(W2)=f (w)
= 12 0.1 + 32 0.2 + … + 92 0.1
= 29.8
(b) / (i) / w / 1 / 3 / 5 / 7 / 9z = (w – 5)2 / 16 / 4 / 0 / 4 / 16
f (w) / 0.1 / 0.2 / 0.4 / 0.2 / 0.1
From the above table, we obtain the probability distribution of Z below.
z / 0 / 4 / 16fZ(z) / 0.4 / 0.4 / 0.2
(ii)E(Z)=
= 0 0.4 + 4 0.4 + 16 0.2
= 4.8
(iii)SinceZ= (W – 5)2
= W 2 – 10W + 25,
E(Z) = E(W 2) – 10E(W) + 25
From the results in (a),
E(Z)= 29.8 – 10 5 + 25
= 4.8
3.(a)E(X) =
(b)E(4X + 3) = 4E(X) + 3 =
(c)E(7 2X) = 7 2E(X) =
4.(a)E(T) == (2) 0.1 + (1) 0.2 + 0 0.3 + 1 0.15 + 2 0.25 = 0.25
(b)E(T2) =
E(T2)= 1.75
(c)E(3T2 + 5) = 3E(T2) + 5 = 3 1.75 + 5 = 10.25
5.(a)f (x) =
f (0) =
f (1) =
f (2) =
f (3) =
(b)E(X) =
(c)E(X2) =
E(X2) [E(X)]2 =
6.(a)= 1
= 1
p + q= ...... (1)
From the given,E(Z)= 4
= 4
q= ...... (2)
Putting (2) into (1),
=
p=
(b)E(Z2 )=
=
= 26.5
E[(Z + 2)2]= E(Z2 + 4Z + 4)
= E(Z2 ) + 4E(Z) + 4
= 26.5 + 4 4 + 4
= 46.5
7.(a)E(X)= 5.5
1 0.1 + a 0.2 + 6 0.3 + b 0.4= 5.5
Thus,a + 2b= 18...... (1)
E(X2 )= 37.3
12 0.1 + a2 0.2 + 62 0.3 + b2 0.4= 37.3
Thus,a2 + 2 b2= 132...... (2)
Solving (1) and (2),
a = 2, b = 8 or a = 10, b = 4
As it is given that ab,
a = 2 and b = 8.
(b)E(X3) = = 271.3
8.(a)E(Y)= 3...... (1)
E[Y(Y 1)]= 12...... (2)
From (2),E(Y2) E(Y)= 12
E(Y2) 3= 12
E(Y2)= 15
(b)E[(Y E[Y])2]= E[(Y 3)2]
= E[Y2 6Y + 9]
= E(Y2) 6E(Y) + 9
= 15 6 3 + 9
= 6
9.(a)The probability distribution of X is:
x / 1 / 2 / 3 / 4 / 5 / 6f (x) / / / / / /
E(X) = = 3.5
(b)E[(X E[X])2]= E[(X 3.5)2]
=
= (or 2.916 7)
10.(a)Since the chip is selected at random, all chips are equally likely to be selected. Hence, we have the following probability distribution for X.
x / 1 / 2 / 3f(x) / / /
(b)E(X)=
=
=
11.(a)
(b)The probability distribution of X is:
x / 0 / 1 / 2 / 3f (x) / / / /
E(X) = = 1.8
12.(a)Let X (in cm) be the diameter of the cake that the girl buys.
The expected diameter= E(X)
= (15 + 18 + 20 + 25 ) cm
= 19.5 cm
(b)The expected top surface area=
=
=
= 309.1 cm2
13.His expected return = $3 000 0.4 + $(1 500) 0.6 = $300
14.(a) The expected value of a $1 bet =
(b)His expected return = = $1 400
15.(a)E(X)=
= 1 0.1 + 2 0.15 + 3 0.4 + 4 0.25 + 5 0.1
= 3.1
(b)The charge $Y = $500X per hour
The expected earning per hour= $E(Y)
= $E(500X)
= $500E(X)
= $500 3.1
= $1 550
16.(a)The expected number of absentees during a working day
= 0 0.80 + 1 0.10 + 2 0.07 + 3 0.03 = 0.33
(b)The expected man-days loss in a year
= 0.33 250 = 82.5
17.Michael’s expected return
= $(51 000 0.1 + 21 000 0.15 + 11 000 0.25 + 1 000 0.5)
= $11 500
18.(a)The sample space is S = {(x, y) : x = 1, 2, 3, 4, 5, 6 and y = 1, 2, 3, 4, 5, 6}.
It is equiprobable, because the dice are fair, and consists of 36 sample points. Hence
P(X = x, Y = y) = .
T = X + Y and can assume values 2, 3, … , 12. If the event {T = t} consists of n sample points of S, then P(T = t) = . For example, {T = 5} consists of the 4 sample points (1, 4), (2, 3), (3, 2), (4, 1) and so P(T = 5) = .
Using this method, the probability distribution of T is obtained and shown below.
t / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12f (t) / / / / / / / / / / /
(b)E(T)=
=
= 7
Note:This result is expected because of the symmetry of the distribution.
(c)The probability distribution of the amount paid by my friend is as follows.
Amount paid, $z / 5 / 10Corresponding sum
of scores, t / <7 / 7
Probability / /
Expected amount per toss paid by my friend= $
= $7
Hence, for the bet to be fair, I should pay $7 to my friend before each toss.
19.The expected claim from the man
= $500 000 (1 0.002 + 0.5 0.015 + 0.25 0.01)
= $6 000
The required premium
= the expected claim + the profit
= $6 000 + $6 000
= $12 000
1
Section 14.4 Variance and Standard Deviation
20.(a)The expected demand
= E(X)
= 10 0.05 + 11 0.1 + 12 0.2 + 13 0.3 + 14 0.2 + 15 0.15
= 12.95 copies
(b)If the news agent orders 13 copies for sale, the probability distribution of his net profit $Y is given in the following table.
x / 10 / 11 / 12 / 13 / 14 / 15no. of copies sold / 10 / 11 / 12 / 13 / 13 / 13
net profit y ($) / 1 / 21 / 43 / 65 / 65 / 65
f (y) [= f (x)] / 0.05 / 0.1 / 0.2 / 0.3 / 0.2 / 0.15
His expected net profit
=
= $(1 0.05 + 21 0.1 + 43 0.2 + 65 0.3 + 65 0.2 + 65 0.15)
= $52.9
(c)If the news agent orders 11 copies for sale, the probability distribution of his net profit $Y is given in the following table.
x / 10 / 11 / 12 / 13 / 14 / 15no. of copies sold / 10 / 11 / 11 / 11 / 11 / 11
y / 33 / 55 / 55 / 55 / 55 / 55
f (y) / 0.05 / 0.1 / 0.2 / 0.3 / 0.2 / 0.15
His expected net profit = $(33 0.05 + 55 0.95) = $53.9
(d)Similar to parts (b) and (c), the expected net profits for different levels of order are calculated and shown in the following.
Order for sale / 10 / 11 / 12 / 13 / 14 / 15Expected net profit ($) / 50 / 53.9 / 55.6 / 52.9 / 43.6 / 29.9
The news agent should order 12 copies of the magazine to maximize his expected net profit.
EXERCISE 14.4 / Section 14.4 Variance and standard deviation(page 315)
1.(a)Mean of X=
= 0 0.6 + 1 0.4
= 0.4
Variance of X=
= (0 – 0.4)2 0.6 + (1 – 0.4)2 0.4
= 0.24
(b)(i)The probability distribution of Y is shown below.
y = 5x – 1 / –1 / 4fY(y) / 0.6 / 0.4
(ii)Mean Y of Y=
= (–1) 0.6 + 4 0.4
= 1
Variance of Y=
= (–1 – 1)2 0.6 + (4 – 1)2 0.4
= 6
(iii)Since Y = 5X – 1,
Mean of Y= 5 (mean of X) – 1
= 5 0.4 – 1,from the result in (a)
= 1
Variance of Y= 52 (variance of X)
= 25 0.24,from the result in (a)
= 6
2.(a)mean = E(X) = = 1 0.2 + 3 0.5 + 8 0.3 = 4.1
E(X2) = = 12 0.2 + 32 0.5 + 82 0.3 = 23.9
Var (X) = E(X2) [E(X)]2 = 23.9 4.12 = 7.09
standard deviation of X = = 2.663
(b)mean of (2X + 3) = E(2X + 3) = 2E(X) + 3 = 2 4.1 + 3 = 11.2
Var (2X + 3) = 22 Var (X) = 4 7.09 = 28.36
standard deviation of (2X + 3) = = 5.325
3.(a)mean= E(Y)
= = 1.875
E(Y2)= = 14.25
Var (Y)= E(Y2) [E(Y)]2
= 14.25 (1.875)2
= 10.7344
= 10.73 (2 d.p.)
standard deviation of Y = = 3.28 (2 d.p.)
(b)mean of (5 4Y) = E(5 4Y) = 4E(Y) = 4 (1.875) = 7.5
Var (5 4Y) = (4)2 Var (Y) = 16 10.734 4 = 171.750 4 = 171.75 (2 d.p.)
standard deviation of (5 4Y) = = 13.11 (2 d.p.)
4.(a) E(3X 5)= 3E(X) 5 = 3 18 5 = 49
Var (3X 5)= 32 Var (X) = 9 4 = 36
(b)== = 1
= = = 1
5.(a)E(2Y + 30) = 2E(Y) + 30 = 2 (10) + 30 = 10
(b)
(c)Var (8 5Y) = 25 Var (Y) = 25 6 = 150
6.(a)X= 1 0.40 + 2 0.35 + 3 0.15 + 4 0.10 = 1.95
X2= = 0.947 5
standard deviation of X = X = 0.973 4
(b)Y= 1 0.10 + 2 0.15 + 3 0.35 + 4 0.40 = 3.05
Y2= = 0.947 5
standard deviation of Y = Y = 0.973 4
(c)The random variables X and Y have the same range and dispersion.
But the distribution of X is skewed to the right while that of Y is skewed to the left.
Hence, XY.
7.(a) U= (3) 0.2 + (1) 0.2 + 0 0.2 + 1 0.2 + 3 0.2 = 0
U2= [] 02
= 4
U= = 2
(b)V= (3) 0.05 + (1) 0.25 + 0 0.40 + 1 0.25 + 3 0.05 = 0
V2= = 1.4
V= = 1.183
(c)The distributions of U and V are symmetric and they have the same mean.
U is uniformly distributed and V is denser in the middle. Hence, the dispersion of U is greater than that of V.
8.The probability distribution of X is:
x / 0 / 1 / 2 / 3 / 4f (x) / / / / /
=
=
=
2= E(X2) 2
=
=
standard deviation = = (or 0.942 8)
9.(a)Since
= 1
(15)= 1
k= 15
(b)E(X)=
=
=
=
=
Var(X)=
=
=
=
10.(a)= 1
= 1
k=
(b)mean= E(T)
=
= (or 3.273)
Var (T)= E(T2) [E(T)]2
=
= 8.926
standard deviation of T = = 2.99
11.(a)From E[(X 1)2]= 9,
we haveE[X2 2X + 1]= 9
E(X2) 2E(X)= 8...... (1)
FromE[(X + 3)2]= 29,
we haveE[X2 + 6X + 9]= 29
E(X2) + 6E(X)= 20...... (2)
Solving (1) and (2), yields
E(X)=
andE(X2)= 11
(b)Var (X) = E(X2) [E(X)]2 = (or 8.75)
(c)Var (3X 7) = 32 Var (X) = (or 78.75)
12.The probability function of X is
mean=
= (0 + 1 + 2 + . . . + 9) 0.1
= 4.5
Var (X)=
= (02 + 12 + 22 + . . . + 92) 0.1 4.52
= 8.25
standard deviation of X == 2.872
13.(a)The probability distribution of X1 is as follows.
Result / Adam wins / Adam losesx1 ($) / 10 / –10
f (x1) / /
(b)E(X1) =
Var(X1) =
(c)The probability distribution of X2 is as follows.
x2 ($) / 100 / –100f (x2) / /
(d)E(X2) =
Var(X2) =
(e)X1 and X2 have equal expected value (= 0), but Var (X2) > Var (X1). When the bet is fair (expected gain = 0), the higher variance reflects a greater variability in the gain and loss and hence a higher degree of risk.
14.The probability function of X is
The probability distribution of X is:
x / 0 / 1 / 2 / 3f (x) / / / /
mean= E(X)
=
= (or 1.667)
Var (X)= E(X 2) [E(X)]2
=
= (or 0.555 6)
15.(a)The probability function of X is
The probability distribution of X is:
x / 0 / 1 / 2 / 3 / 4f (x) / / / / /
(b)mean= E(X)
=
= 2
Var (X)=
=
= 1
standard deviation of X = = 1
16.(a)The probability function of X is:
The probability distribution of X is:
x / 0 / 1 / 2 / 3 / 4f (x) / 0.281 7 / 0.469 6 / 0.216 7 / 0.031 0 / 0.001 0
(b)mean= E(X)
= 0 0.281 7 + 1 0.469 6 + 2 0.216 7 + 3 0.031 0 + 4 0.001 0
= 1
E(X2)= 02 0.281 7 + 12 0.469 6 + 22 0.216 7 + 32 0.031 0 + 42 0.001 0
= 1.631 4
Var (X)= 1.631 4 12
= 0.631 4
17.(a)E(Z) =
Var (Z) = = 1
(b)(i)The mean =
= 10 0.1 + 12 0.3 + 14 0.25 + 16 0.2 + 18 0.15
= 14
The variance 2=
= (10 – 14)2 0.1 + (12 – 14)2 0.3 + + (18 – 14)2 0.15
= 6
(ii)The standardized random variable can assume the following values:
i.e.
or, equivalently,
18.Y= aX + b
E(Y)= E(aX + b)
= a E(X) + b
45= 30a + b...... (1)
Var (Y)= Var (aX + b)
= a2 Var (X)
20= 5a2...... (2)
From (2),a = 2 or 2...... (3)
Putting (3) into (1),
when a = 2,45= 30(2) + b
b= 15
when a = 2,45= 30(2) + b
b= 105
(a = 2, b = 15) or (a = 2, b = 105).
19.(a)Mean X of X = 38, by symmetry of the probability distribution
Variance =
= (36 – 38)2 0.1 + (37 – 38)2 0.2 + + (40 – 38)2 0.1
= 1.2
(b)Mean Y of Y=
= 36 0.6 + 37 0.1 + + 40 0.1
= 37
Variance of Y=
= (36 – 37)2 0.6 + (37 – 37)2 0.1 + + (40 – 37)2 0.1
= 2
(c)Both results show a tendency of the respondents to give lower ages. The second question gives an even greater downward bias; it also gives a greater variability (greater variance) in the answers.
(d)Question 1 should be used as it gives a smaller bias and a lower variability in the answers, and hence more reliable answers in general.
1
Revision Exercise 14
(e)A better question would be
“What is your date of birth?”
This avoids the direct mention of the age or the word “old”.
20.(a)The 3 notes drawn by the lady may consist of 0, 1, 2 or 3 $500 notes, corresponding to the values of 300, 700, 1 100 and 1 500, respectively, of X.
The probability of drawing k $500 notes and (3 – k) $100 notes is
Hence, we have the following probability distribution for X.
k / 0 / 1 / 2 / 3x / 300 / 700 / 1 100 / 1 500
f (x) / / / /
(b)Expected amount= $E(X)
=
= $700
(c)Variance of X=
=
= 87 272 (in dollar2)
(d)P(X > 1 000)= P(X = 1 100) + P(X = 1 500)
=
=
(e)For both ladies to get more than $1 000, the only possible case is when each gets $1 100. Then each must get 2 $500 notes and 1 $100 note.
The probability=
=
REVISION EXERCISE 14 / Revision exercise 14(page 320)
1.(a)P(1 T 1)= P(T = 0) + P(T = 1)
= 0.3 + 0.4
= 0.7
(b)E(T)= 2 0.15 + (1) 0.1 + 0 0.3 + 1 0.4 + 2 0.05
= 0.1
E(2T + 3)= 2E(T) + 3
= 2 0.1 + 3
= 3.2
2.(a) = 1
= 1
k=
(b)P(Y 4)= P(Y = 5) + P(Y = 6)
=
=
(c)mean=
=
=
E(Y2)=
=
=
Variance of Y= E(Y2) mean2
=
= (or 3.061 5)
3.(a)Mean X of X=
= 1 0.1 + 2 0.25 + 4 0.3 + 8 0.25 + 16 0.1
= 5.4
Varianceof X=
= 12 0.1 + 22 0.25 + 42 0.3 + 82 0.25 + 162 0.1 – 5.42
= 18.34
(b)The probability distribution of Y is given below.
y = ln x / 0 / ln 2 / 2 ln 2 / 3 ln 2 / 4 ln 2fY(y) / 0.1 / 0.25 / 0.3 / 0.25 / 0.1
(c)Mean Y of Y=
= 0 0.1 + (ln 2) 0.25 + (2 ln 2) 0.3 + (3 ln 2) 0.25 + (4 ln 2) 0.1
= 2 ln 2
Variance of Y=
= (–2 ln 2)2 0.1 + (–ln 2)2 0.25 + 02 0.3 + (ln 2)2 0.25
+ (2 ln 2)2 0.1
= 1.3 (ln 2)2
(d)Probability distribution of X:
Probability distribution of Y:
(e)The probability distribution of X is positively skewed;
the probability distribution of Y is symmetric about y = 2 ln 2.
X has a higher mean (X = 5.4) than Y (Y = 2 ln 2 = 1.386 3).
X has a much larger variance ( = 18.34) than Y ( = 0.624 6).
In fact, Y is a logarithmic transformation of X and has the effect of reducing the skewness, the mean and the variance of X. In this particular question, the transformation makes the distribution symmetrical.
4.(a)The expected percentage of loss
=
= (0 0.80 + 10 0.11 + 25 0.05 + 50 0.03 + 100 0.01)%
= 4.85%
Expected loss= $300 000 4.85%
= $14 550
Annual premium per policy= $(14 550 + 2 000)
= $16 550
(b)(i)P(all 3 cars will claim)= P(all 3 cars will have certain loss)
= (1 – 0.8)3
= 0.008
(ii)P(at least one of the 3 cars will claim)
= 1 – P (no car will claim)
= 1 – 0.83
= 0.488
5.(a)
From the above figure, the probability distribution of T is
t / 0 / 2 / 4 / 5 / 7 / 10P(T = t) / / / / / /
(b)Expected return of Donna
= $
= $
6.[Public Examination question]
(a) (b) $85 600
7.[Public Examination question]
(a)(i) 0.9; 0.4; 0.36 (ii) Yes (b) 3.88
8.[Public Examination question]
(a) (c)
9.(a)E(X) = = 5.1
0(0.1) + 1(0.2) + 4(0.15) + 6(0.3) + n(0.25)= 5.1
n= 10
(b)E(X2)=
= 02 0.1 + 12 0.2 + 42 0.15 + 62 0.3 + 102 0.25
= 38.4
Variance of X= E(X2) [E(X)]2
= 38.4 5.12
= 12.39
(c)P(2X + 1 5)= P(X 2)
= 0.15 + 0.3 + 0.25
= 0.7
10.(a)E(X)=
= 1
=
= 2
(b)(i)
From the above figure, the probability distribution of Y is
y / 6 / 3 / 0 / 3 / 6P(Y = y) / / / / /
(ii)E(Y)=
= 2
E(Y 2)=
= 14
Var(Y)= E(Y 2) [E(Y)]2 = 14 (2)2 = 10
Hence, Var(2Y)= 22 Var(Y) = 40
E(Y + 1)= E(Y) + 1 = 1
11.(a)(i)P(3 bottles) =
(ii)P(2 bottles)
= P(HHM) + P(HMH) + P(MHH),where H-hit, M-miss
=
= (or 0.169 0)
(b)The expected value= $
= – $ (or – $0.430 4)
12.(a)
(i)P(undistorted and finished satisfactorily)= 0.9 0.92
= 0.828
(ii)P(distorted and not finished satisfactorily)= 0.1 0.25
= 0.025
(b)The expected profit
= $[100 0.828 + (30) 0.025 + 60 (1 0.828 0.025)] 200
= $18 174
13.(a)Mean of X=
= 0 0.01 + 1 0.02 + 2 0.10 + 3 0.26 + 4 0.38 + 5 0.23
= 3.67
Variance of X=
= 02 0.01 + 12 0.02 + + 52 0.23 – 3.672
= 1.121 1
(b)(i)P(the pill will have effect on not more than 3 patients for one group)
= P(X 3)
= 1 – P(X = 4) – P(X = 5)
= 1 – 0.38 – 0.23
= 0.39
P(the pill will have effect on not more than 3 patients for each group)
= (0.39)4
= 0.023 1
(ii)For each of the 4 groups, the number of patients brought under control is not more than 3. Hence, by (i) above, the probability of the results = 0.023 1. This is a small probability (< 0.05), indicating that the observed results constitute a rare event if the test report is valid. Hence, the observation does not support the test report.
14.(a)
= 5 0.07 + 6 0.16 + + 11 0.03
= 7.71
= (52 0.07 + 62 0.16 + + 112 0.03) – 7.712
= 2.285 9
(b)
= 2 0.10 + 3 0.11 + + 9 0.04
= 5.26
= (22 0.10 + 32 0.11 + + 92 0.04) – 5.262
= 3.632 4
(c)Since , the mixed text and colour graphics type has a greater variation in print speed.
(d)Some causes are:
1.document complexity
2.software program
3.printing mode
4.processing speed of the computer
(e)
= P(7.71 – 1.511 9 < X < 7.71 + 1.511 9)
= P(6.198 1 < X < 9.221 9)
= P(X = 7) + P(X = 8) + P(X = 9)
= 0.24 + 0.21 + 0.19
= 0.64
(f)
= P(5.26 – 1.905 9 < Y < 5.26 + 1.905 9)
= P(3.354 1 < Y < 7.165 9)
= P(Y = 4) + P(Y = 5) + P(Y = 6) + P(Y = 7)
= 0.13 + 0.20 + 0.18 + 0.15
= 0.66
(g)Yes, acceptable. In an office, the printer may be used in the normal mode. The results in (e) and (f) indicate that there are reasonable probabilities for the print speeds claimed in the brochure to be achieved.
(h)
The conditions are represented in the above tree diagram, where the conditional probabilities for the print speeds have been derived from the given probability distributions of X and Y.
P(black text | print speed 7 ppm)
=
=
= 0.846 2
15.(a)Number of ways the 4 customers can arrange their seats
= (4 – 1)! = 6 / Circular permutation(b)(i)When the weather is fine, the expected number of customers
= 0 0.10 + 1 0.15 + 2 0.20 + 3 0.25 + 4 0.30
= 2.5
Expected earning from the table= $100 2.5
= $250
(ii)When the weather is not fine, the expected earning from the table
= $100(0 0.30 + 1 0.25 + 2 0.25 + 3 0.15 + 4 0.05)
= $140
(c)Expected earning from the table on that day
= P(the weather is fine) (expected earning when it is fine)
+ P(the weather is not fine) (expected earning when it is not fine)
= $(0.80 250 + 0.20 140), using the results in (b)(i) and (ii)
= $228
(d)(i)Expected earning from the table for that month (30 days)
= $
= $6 180
(ii)
The above tree diagram shows the information relevant to the calculation of the probability required.
P(the weather is fine | the table is occupied by 2 customers)
=
=
=
16.[Public Examination question]
(a)(i) (ii) (b) 3; 1.060 7
17.[Public Examination question]
(a) / No. of cars kept / 1 / 2 / 3 / 4 / 5 / 6Expected daily profit ($) / 132.5 / 252.6 / 341.5 / 378.4 / 350.4 / 257.5
optimum = 4
(b)$378.4; $308.9
18.[Public Examination question]
(b) (c)(i) 0.090 2 (ii) 0.685 1 (iii) 10
19.[Public Examination question]
(a) (c) (d) $54 500
20.[Public Examination question]
(a) (b)(ii) (iii)
1