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COM2023 2010-11, Exercises Chapters 2-4
Data
1. Find the mean, median and mode of the numbers 3,5,2,6,5,5,9,2,8,6.
2. Find the means and the variation of the sample 12,6,7,3,15,10,18,5.
3. Find the mean and standard deviations of the following data:
length (cm) midpoint frequency
60 − 63
61.5
5
63 − 66
64.5
18
67.5
42
66 − 69
69 − 72
70.5
27
72 − 75
73.5
8
Probability
4. Find the probability of a 4 turning up at least once in two tosses of a fair die by using an
appropriate addition law.
5. Bag A contains 4 white balls and 2 black balls, whilst bag B contains 3 white balls and 5
black balls. One ball is drawn from each bag. Find the probability that
(a) both balls are white;
(b) both balls are black.
6. Three balls are drawn successively from a box containing 6 red, 4 white and 5 blue balls.
Find the probability that they are drawn in the order red, white and blue if before the next
ball is drawn each ball is
(a) replaced;
(b) not replaced.
7. A batch consists of l00 valves, 5 of which are faulty. A random sample of 4 valves is chosen.
What is the probability that it contains one faulty valve?
Distributions
The exercises 13-17 are based on section 4.11.
8. Find the probability that in five tosses of a fair die a 3 appears
(a) at no time;
(b) three times;
(c) five times.
9. If 20 percent of the bolts produced by a machine are defective, find the probability that out
of 4 bolts chosen at random
(a) exactly one is defective;
(b) at most two are defective.
10. Ten percent of the tools produced in a certain factory are defective. Find the probability that
in a sample of 10 tools chosen at random, exactly two will be defective using
1
(a) the binomial distribution;
(b) the Poisson approximation.
11. A bag contains 2 white balls and 3 black balls. Four persons A, B, C, D each draw a ball in
that order and do not replace it. The first person to draw a white ball receives 10 pounds.
Find the expectations of each person.
12. A random variable X satisfies a normal distributions with mean 4 and variance 9 and a random
variable Y satisfies a normal distributions with mean 6 and variance 5, i.e., X ∼ N (4, 9) and
Y ∼ N (6, 5).
(a) What Matlab-commands would you use to find P (X < 3), P (X > 7) and P (X < 5 |
X > 0)?
(b) What Matlab-commands would you use to find P (Y ≤ 8), P (Y > 15) and P (4 < Y <
10)?
(c) What are the expectation and variance of V = X + 2Y − 6 and W = 3X − 4Y ?
(d) What normal distributions are satisfied by V = X + 2Y − 6 resp. W = 3X − 4Y ?
13. The random variable X satisfies a Bernoulli distribution, namely X ∼ B(15, 0.4). Find the
following probabilities:
P (X = 8), P (X = 11), P (X ≤ 2) and P (X < 14).
14. The random variables U and V satisfy Poisson distributions, namely U ∼ P o(7) and V ∼
P o(3). Find the following probabilities:
P (U = 4), P (V = 5), P (U < 2), P (U + V ) = 9 and P (U + V ≥ 2).
15. The random variable W satisfies a uniform distribution, namely W ∼ U [10, 30]. Find the
following probabilities:
P (W > 24) and P (W < 35).
16. A company produces telephones and 15% of the telephones need to be replaced while under
warranty. A shop sells 12 of those phones in a one month period. What is the probability
that exactly 3 of these phones will be replaced under warranty?
17. Small aircraft arrive at an airport a rate of 8 an hour.
(a) What is the probability that exactly 10 small aircraft arrive during a one hour period?
(b) What is the probability that less than 2 small aircraft arrive during a one hour period?
(c) What assumption did you make to answer (a) and (b)?
2
Solutions
1. mean = 5.1, median = 5, mode = 5
In ascending order the numbers are 2,2,3,5,5,5,6,6,8,9. The mean is 2+2+3+5+5+5+6+6+8+9
=
10
5.1 . The median is the arithmetic mean of the two centrally occurring numbers, i.e 5+5
=
5.
2
The mode is the most frequently occurring number, i.e, this is 5.
2. mean is µ =
76
8
and variance is s2 = 27.14
We have
X
n = 8,
Hence the mean is
s2 =
1
7
X
xi = 12 + 5 + . . . = 76,
76
8
912 −
x2i = 912.
and the sample variance is
762
8
= 27.14.
Or we can use the definition of the variance. The sequence xi − x is
2.5, −3.5, −2.5, −6.5, 5.5, 0.5, 8.5, −4.5
Thus
P
(xi − x)2 = 190 and the sample variance is s2 =
190
7
= 27.14.
3. mean is µ = 67.95 and standard deviation is s = 2.93
We have
n=
X
fi = 100,
Thus mean is
1
s =
99
2
6795
100
X
X
fi xi = 6795,
fi x2i = 462573.
and the sample variance is
67952
462573 −
= 8.61.
100
And the standard deviation is s = 2.93.
4. probability is
11
36
Let A and B be the events of a 4 appearing on the first respectively second tosses. Then
P (A) = P (B) = 16 and the event A ∩ P represents the event of a 4 turning up on both tosses,
1
so that P (A ∩ B) = 36
. Thus we have that
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
5. (a) probability is
2
1
− .
6 36
1
4
4
The probability of a white ball from bag A is P (W1 ) = 4+2
= 32 . Also, the probability
3
3
of a white ball from bag B is P (W2 ) = 3+5 = 8 . The events are independent, hence the
probability of both evens occurring is P (W1 ) · P (W2 ).
(b) probability is
5
24
The probability of both being black is
3
2
6
·
5
8
=
5
24 .
6. (a) probability is
8
225
Let RW B be a red ball on the first draw, a white ball on the second draw and a blue
ball on the third draw. If each ball is replaced then events are independent. The total
6 4 5
8
number of balls is 15, hence P (RW B) = 15
15 15 = 225 .
(b) probability is
4
91
If each ball is not replaced, then the events are not independent. Thus on the second
draw we need the probability that a white ball is drawn given that the first drawing
4
produced a red ball. This is P (W | R) = 14
. Continuing like this we get
P (RW B) = P (R) P (W | R) P (B | RW ) =
6 4 5
4
= .
15 14 13
91
7. probability is 0.176
5
The probability of a valve being faulty is 100
=
sample is faulty and the remainder are correct is
1
20 .
The probability that the first in the
5 95 94 93
·
·
· .
100 99 98 97
(After taking the faulty one, there are 99 valves left, 95 of which are not faulty, etcetera).
Next we look at the probability that the second valve in the sample is faulty, this is
95 5 94 93
·
·
· ,
100 99 98 97
which is equal to the first valve being faulty. In the same way we determine that probability
that the third and the fourth valve are faulty is also the same as the previous two expressions.
Altogether this gives for the total probability of one faulty valve in the sample
4·
5 · 95 · 94 · 93
= 0.176.
100 · 99 · 98 · 97
8. (a) the probability is
3125
7776
Define the random variable X to be the number of 3’s in the experiment. The probability
of a 3 in a single toss is 61 . The probability of no 3 in a single toss is 56 . Using the binomial
distribution we get that
0 5
5
1
5
3125
P (X = 0) =
=
.
0
6
6
7776
(b) the probability is
125
3888
Using the binomial distribution we get that
3 2
5
1
5
125
P (X = 3) =
=
.
6
6
3888
3
(c) the probability is
1
7776
Using the binomial distribution we get that
5 0
5
1
5
1
P (X = 5) =
=
.
5
6
6
7776
9. (a) the probability is 0.4096
The probability of a defective bolt is p = 0.2. Using the binomial
the
distribution,
4 1
44
3
probability of exactly one defective bolt out of 4 is P (X = 1) = 1 p (1 − p) = 54 .
4
(b) the probability is 0.9728
The probability of at most two defective bolts is
4 3
4 4
1
1 − P (X = 4) − P (X = 3) = 1 −
p (1 − p) −
p (1 − p)0 = 0.9728.
3
4
10. (a) the probability is 0.1937
The probability of a defective tool is p = 0.1. We take a selection of n = 10. With the
binomial distribution we get that
10 2
10 · 9
(0.1)2 (0.9)8 = 0.1937.
P (X = 2) =
p (1 − p)8 =
2·1
2
(b) the probability is 0.1839
We have µ = np = 1, hence with the Poisson distribution we get that P (X = 2) =
e−1 12
1
= 2e
.
2!
11. E(A) = £4, E(B) = £3, E(C) = £2, E(D) = £1
The probability that A wins is P (A wins) =
pounds.
2
2+3 .
The probability of A loses followed by B wins is
3
10 · 10 = 3 pounds.
Similarly C’s expectation is
322
543
Hence the expectation of A is
32
54
=
3
10 .
2
5
· 10 = 4
Hence the expectation of B is
· 10 = 2. And D’s expectation is
3212
5432
· 10 = 1.
Note that as there are only 3 black balls present, one person will win.
12. (a) The Matlab-commands are:
i. P (X < 3) by normcdf(3,4,3),
ii. P (X > 7) by 1-normcdf(7,4,3),
iii. P (X < 5 | X > 0) by (normcdf(5,4,3)-normcdf(0,4,3))/(1-normcdf(0,4,3)).
(b) The Matlab-commands are:
i. P (Y ≤ 8) by normcdf(8,6,sqrt(5)),
ii. P (Y > 15) by 1-cdfnorm(15,6,sqrt(5)),
iii. P (4 < Y < 10) by normcdf(10,6,sqrt(5))-normcdf(4,6,sqrt(5)).
(c) E(V ) = 10, Var(V ) = 29 and E(W ) = −12 and Var(W ) = 161
Using the rules for expectations and variance, we get
E(V ) = E(X + 2Y − 6) = E(X) + 2E(Y ) − 6 = 4 + 12 − 6 = 10
Var(V ) = Var(X + 2Y − 6) = 12 Var(X) + 22 Var(Y ) = 9 + 20 = 29.
E(W ) = E(3X − 4Y ) = 3E(X) − 4E(Y ) = −12
Var(W ) = Var(3X − 4Y ) = 32 Var(X) + (−4)2 Var(Y ) = 81 + 80 = 161.
(d) Using that sums of normal distributions give normal distributions, again, we get with
the results above that V ∼ N (10, 29) and W ∼ N ().
5
13.
14.
17.
6
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