Question

Google it: According to a recent report, 65% of Internet searches in a particular month used the Google search engine. Assume that a sample of 20 searches is studied. Round the answers to four decimal places. Part 1 of 4 (a) What is the probability that exactly 15 of them used Google? The probability that exactly 15 of them used Google is Part 2 of 4 (b) What is the probability that 10 or fewer used Google? The probability that 10 or fewer used Google is

Answer 1

Let ,

X + Bn = 20, p = 0.65)

Therefore , the pmf of X is ,

$P(X=x)=\binom{n}{x}p^xq^{n-x}$ ; x=0,1,2,.......,n and q=1-p

= 0 ; otherwise

The table given below ,

X	$\binom{n}{x}$	$p^x$	$q^{n-x}$	P(X=x)
0	1	1	7.60958E-10	8E-10
1	20	0.65	2.17417E-09	3E-08
2	190	0.4225	6.2119E-09	5E-07
3	1140	0.274625	1.77483E-08	6E-06
4	4845	0.178506	5.07094E-08	4E-05
5	15504	0.116029	1.44884E-07	0.0003
6	38760	0.075419	4.13955E-07	0.0012
7	77520	0.049022	1.18273E-06	0.0045
8	125970	0.031864	3.37922E-06	0.0136
9	167960	0.020712	9.65492E-06	0.0336
10	184756	0.013463	2.75855E-05	0.0686
11	167960	0.008751	7.88156E-05	0.1158
12	125970	0.005688	0.000225188	0.1614
13	77520	0.003697	0.000643393	0.1844
14	38760	0.002403	0.001838266	0.1712
15	15504	0.001562	0.005252188	0.1272
16	4845	0.001015	0.01500625	0.0738
17	1140	0.00066	0.042875	0.0323
18	190	0.000429	0.1225	0.01
19	20	0.000279	0.35	0.002
20	1	0.000181	1	0.0002

Part 1 of 4 : P(X=15)=0.1272

Part 2 of 4 :

$P(X\leq 10)=P(X=10)+P(X=9)+P(X=8)+P(X=7)+P(X=6)+P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)$

=0.0686+0.0336+0.0136+0.0045+0.0012+0.0003+0+0+0+0+0

=0.1218

Part 3 of 4 :

$P(X>15)=P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)$

=0.0738+0.0323+0.01+0.002+0.0002

=0.1183

Part 4 of 4 :

$P(X<8)=P(X=7)+P(X=6)+P(X=5)+P(X=4)+P(X=3)+P(X=2)+P(X=1)+P(X=0)$

=0.0045+0.0012+0.0003+0+0+0+0+0

=0.0060

Since , P(X<8)=0.0060 <0.05

Therefore , it should be unusual.