Question

5.   According to a reliable source, 65% of murders are committed with a firearm. Suppose 15 murders are randomly selected. First construct a relative and cumulative frequency distribution for the situation. Then confirm that it is both a probability and binomial probability distribution.

a.   Compute the mean.
b.   Compute the standard deviation.
c.   Would a sample of 15 with 6 murders committed with a firearm be considered unusual? Justify your reasoning.
d.   Find the probability that exactly 10 murders are committed with a firearm.
e.   Find the probability that at most 11 murders are committed with a firearm.
f.   Find the probability that at least 12 murders are committed with a firearm
i.   Find the probability that between 9 and 13 murders are committed with a firearm.

Answer 1

Probability that a murder is committed with firm arm, p = 0.65

n = 15

Relative and cumulative frequency distribution:

X	P(X) = BINOM.DIST(x, 15, 0.65, 0)	P(X<=x) = BINOM.DIST(x, 15, 0.65, 1)
0	0.0000	0.0000
1	0.0000	0.0000
2	0.0001	0.0001
3	0.0004	0.0005
4	0.0024	0.0028
5	0.0096	0.0124
6	0.0298	0.0422
7	0.0710	0.1132
8	0.1319	0.2452
9	0.1906	0.4357
10	0.2123	0.6481
11	0.1792	0.8273
12	0.1110	0.9383
13	0.0476	0.9858
14	0.0126	0.9984
15	0.0016	1.0000

a) Mean, µ = n*p = 9.75

b) Standard deviation, σ = √(n*p*(1-p)) = 1.8473

c) P(X=6) = BINOM.DIST(6, 15, 0.65, 0) = 0.0298

As the probability is less than 0.05. so we can say that it is unusual.

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d) P(X=10) = BINOM.DIST(10, 15, 0.65, 0) = 0.2123

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e) P(X<=11) =

from table probability is

= 0.8273

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f) P(X>=12) = 1- P(X<12)

= 1 - P(X<=11)

= 1 - 0.8273 = 0.1727

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i) P(9< X<13) = P(X<13) -P(X<=9)

= P(X<= 12) -P(X<=9)

= 0.9383 - 0.4357 = 0.5026