Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 17
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20®
Answers
1. Binomial distribution P(X = r) = C p (1 p)
where p is the probability of single ‘success’
which is ‘tyre burst’.
17
1
16
(a) P(X = 1) =
C
(0.05)
(0.95)
= 0.3741
1
(b)
P(X
3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
17
16
17
16
2
15
= (0.95)
+ 17(0.05)(0.95)
+
(0.05)
(0.95)
2
1
17
16
15
3
14
+
(0.05)
(0.95)
= 0.9912
3
2
1
17
16
(c) P(X
2) = 1
P[(X = 0)
(X = 1)] = 1
(0.95)
17(0.05)(0.95)
= 0.2077
2.
(a) P (distortion) = 0.01 for each digit. This is a binomial situation in which the probability
of ‘success’ is 0.01 = p and there are n = 8 trials.
A word is decoded incorrectly if there are two or more digits in error
P(X
2) = 1
P[(X = 0)
(X = 1)]
8
8
8
7
= 1
C
(0.99)
C
(0.01)(0.99)
= 0.00269
0
1
(b) Same as (a) with n = 10. Correct decoding if X
2
P(X
2) = P[(X = 0)
(X = 1)
(X = 2)]
10
9
2
8
= (0.99)
+ 10(0.01)(0.99)
+ 45(0.01)
(0.99)
= 0.99989
3. Let X be a random variable ‘number of answers guessed correctly’ then for each question
1
(i.e. trial) the probability of a ‘success’ =
. It is clear that X follows a binomial distribution
5
with n = 10 and p = 0.2.
1
P (randomly choosing correct answer) =
n = 10
5
P(8 or more correct) = P[(X = 8)
(X = 9)
(X = 10)]
10
8
2
10
9
10
10
=
C
(0.2)
(0.8)
+
C
(0.2)
(0.8) +
C
(0.2)
= 0.000078
8
9
10
4. (a) 0.9841 (b) 0.9999 (c) 1.0000
5. P(X
38) = P(X = 38) + P(X = 39) + P(X = 40) = 0.00626 + 0.1067 + 0.88676 = 0.99975
6. (a) 0.3487 (b) 0.3874 (c) 0.1937 (d) 0.0702
7. 0.15 (total defectives = 0 + 34 + 32 + 9 + 0 out of 500 tested); 44, 39, 14, 2, 0, 0
8. (a) 0.0016, 0.0256, 0.1536, 0.4096, 0.4096;
(b)(i) 0.9728 (b)(ii) 0.9988
33
HELM (2008):
Section 37.2: The Binomial Distribution
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education