Question

According to a​ survey, 63​% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed last year are randomly​ selected, and the number cleared by arrest or exceptional means is recorded. When technology is​ used, use the Tech Help button for further assistance.

​(a) Find the probability that exactly 40 of the murders were cleared. ​

(b) Find the probability that between 35 and 37 of the​ murders, inclusive, were cleared.

​(c) Would it be unusual if fewer than 18 of the murders were​ cleared? Why or why​ not?

Answer 1

We are given p = 0.63

Let X be the number of murders cleared by arrest or exceptional means.

Then X ~ Binomial(50,0.63)

(i) P(X=40) = 0.0046484267

(ii) P(35<=X<=37) = 0.1541893629

(iii) P(X<18) = 0.0000300606

Yes it would be unusual since the probability of that happening is only 0.0000300606

Using excel, the probability table:

X	P(X=x)
0	0.000000000000
1	0.000000000000
2	0.000000000000
3	0.000000000000
4	0.000000000000
5	0.000000000000
6	0.000000000000
7	0.000000000000
8	0.000000000000
9	0.000000000100
10	0.000000000500
11	0.000000003300
12	0.000000018500
13	0.000000092200
14	0.000000415100
15	0.000001696300
16	0.000006318300
17	0.000021516300
18	0.000067165600
19	0.000192611600
20	0.000508338400
21	0.001236498700
22	0.002775286500
23	0.005752767700
24	0.011019659800
25	0.019513732700
26	0.031948159100
27	0.048353970600
28	0.067630215600
29	0.087358321400
30	0.104121674900
31	0.114379520800
32	0.115635377100
33	0.107396001300
34	0.091431730800
35	0.071168482400
36	0.050491153000
37	0.032529727500
38	0.018948682000
39	0.009927376000
40	0.004648426700
41	0.001930460700
42	0.000704357300
43	0.000223127600
44	0.000060441800
45	0.000013721900
46	0.000002539600
47	0.000000368000
48	0.000000039200
49	0.000000002700
50	0.000000000100

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