Question

57% of pet owners greet their pet before spouse of children. If 20 pet owners are randomly selected, find the probability that... Round to 4 decimals. fewer than 12 pet owners greet their pet first? at least 8 owners will greet their pet first? more than 15 owners will greet their pet first? exactly 10 greet their pet first? 5 or less pet owners greet their pet first?

Answer 1

Solution-:

Let, X- Number of pet owners greet their pet before spouse of children

Given:

This situation is suitable for binomial distribution with parameter

The p.m.f. of X is given by,

  

  

By using MS-Excel we prepare dollowing table:

	n=20	p=0.57	q=0.43
x	nCx	p^x	q^(n-x)	P(X=x)
0	1	1	4.67056E-08	0.0000
1	20	0.57	1.08618E-07	0.0000
2	190	0.3249	2.52599E-07	0.0000
3	1140	0.185193	5.8744E-07	0.0001
4	4845	0.10556	1.36614E-06	0.0007
5	15504	0.060169	3.17707E-06	0.0030
6	38760	0.034296	7.38854E-06	0.0098
7	77520	0.019549	1.71826E-05	0.0260
8	125970	0.011143	3.99596E-05	0.0561
9	167960	0.006351	9.29294E-05	0.0991
10	184756	0.00362	0.000216115	0.1446
11	167960	0.002064	0.000502593	0.1742
12	125970	0.001176	0.00116882	0.1732
13	77520	0.00067	0.002718186	0.1413
14	38760	0.000382	0.006321363	0.0936
15	15504	0.000218	0.014700844	0.0496
16	4845	0.000124	0.03418801	0.0206
17	1140	7.08E-05	0.079507	0.0064
18	190	4.03E-05	0.1849	0.0014
19	20	2.3E-05	0.43	0.0002
20	1	1.31E-05	1	0.0000

We find,

1) P[Fewer than 12 pet owners greet their pet first]  

  

The required probability is 0.5136.

2) P[at least 8 owners will greet their pet first]

  

  

   The required probability is 0.0397

3) P[more than 15 owners will greet their pet first]

  

  

The required probability is 0.0286

4) P[ exactly 10 greet their pet first]  

5) P[5 or less pet owners greet their pet first]

  

  

The required probability is 0.0038.