Chapter 6.2, Problem 31E

Google it: According to a report of the Nielsen Company, 76% of Internet searches used the Google search engine. Assume that a sample of 25 searches is studied.

What is the probability that exactly 20 of them used Google?

What is the probability that 15 or fewer used Google?

What is the probability that more than 20 of them used Google?

Would it be unusual if fewer than 12 used Google?

To determine

To find: the probability of using Google exactly 20

Answer to Problem 31E

The probability of using Google exactly 20 is $0.17485$

Explanation of Solution

Given:

Users of Google = 76%

Sample study = 25

Formula used:

Probability of binomial distribution is calculated as

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ \because q=1-p\end{array}$

Calculation:

Here, $p=0.76$

  $n=25$ and $x=20$

  $q=1-p=1-0.76=0.24$

Probability of binomial distribution is,

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ $ P(20)=C{}_{20}{(0.76)}^{20}{(0.24)}^{25-20}$$ P(20)=\left(53130\right){(0.76)}^{20}{(0.24)}^{25-20}$P(20)=0.17485\end{array}$

Conclusion:

Therefore, the probability of using Google exactly 20is $0.17485$

To determine

To find: the probability of using Google 15 or fewer

Answer to Problem 31E

The probability of using Google 15 or feweris $0.05587$

Explanation of Solution

Given:

Users of Google = 76%

Sample study = 25

Formula used:

Probability of binomial distribution is calculated as

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ \because q=1-p\end{array}$

Calculation:

Here, $p=0.76$

  $n=25$ and $x\le 15$

  $q=1-p=1-0.76=0.24$

  $\begin{array}{l}P(\text{ 15 or fewer})=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+\\ P(10)+P(11)+P(12)+P(13)+P(14)+P(15)\end{array}$

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(0)=C{}_0{(0.76)}^0{(0.24)}^{25-0}$

  $P(0)=0.32009\times {10}^{-15}$

  $P(1)=C{}_1{(0.76)}^1{(0.24)}^{25-1}$

  $P(1)=2.534\times {10}^{-14}$

  $P(2)=C{}_2{(0.76)}^2{(0.24)}^{25-2}$

  $P(2)=9.629\times {10}^{-13}$

  $P(3)=C{}_3{(0.76)}^3{(0.24)}^{25-3}$

  $P(3)=2.3378\times {10}^{-11}$

  $P(4)=C{}_4{(0.76)}^4{(0.24)}^{25-4}$

  $P(4)=4.07\times {10}^{-10}$

  $P(5)=C{}_5{(0.76)}^5{(0.24)}^{25-5}$

  $P(5)=5.415\times {10}^{-9}$

  $P(6)=C{}_6{(0.76)}^6{(0.24)}^{25-6}$

  $P(6)=5.716\times {10}^{-8}$

  $P(7)=C{}_7{(0.76)}^7{(0.24)}^{25-7}$

  $P(7)=4.913\times {10}^{-7}$

  $P(8)=C{}_8{(0.76)}^8{(0.24)}^{25-8}$

  $P(8)=3.5\times {10}^{-6}$

  $P(9)=C{}_9{(0.76)}^9{(0.24)}^{25-9}$

  $P(9)=0.0000209$

  $P(10)=C{}_{10}{(0.76)}^{10}{(0.24)}^{25-10}$

  $P(10)=0.00010$

  $P(11)=C{}_{11}{(0.76)}^{11}{(0.24)}^{25-11}$

  $P(11)=0.000458$

  $P(12)=C{}_{12}{(0.76)}^{12}{(0.24)}^{25-12}$

  $P(12)=0.00169$

  $P(13)=C{}_{13}{(0.76)}^{13}{(0.24)}^{25-13}$

  $P(13)=0.00535$

  $P(14)=C{}_{14}{(0.76)}^{14}{(0.24)}^{25-14}$

  $P(14)=0.0145$

  $P(15)=C{}_{15}{(0.76)}^{15}{(0.24)}^{25-15}$

  $P(15)=0.03378$

  $\begin{array}{l}P(\text{ 15 or fewer})=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+\\ P(10)+P(11)+P(12)+P(13)+P(14)+P(15)\\ =0.32009\times {10}^{-15}+2.534\times {10}^{-14}+9.629\times {10}^{-13}+2.3378\times {10}^{-11}+4.07\times {10}^{-10}+5.415\times {10}^{-9}+5.716\times {10}^{-8}\\ 4.913\times {10}^{-7}+3.5\times {10}^{-6}+0.0000209+0.00010+0.000458+0.00169+0.00535+0.0145+0.03378\\ =0.05587\\ \end{array}$

Conclusion:

Therefore, the probability of using Google 15 or fewer is $0.05587$

To determine

To find: the probability of using Google more than 20

Answer to Problem 31E

The probability of using Google more than 20is $0.24836$

Explanation of Solution

Given:

Users of Google = 76%

Sample study = 25

Formula used:

Probability of binomial distribution is calculated as

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ \because q=1-p\end{array}$

Calculation:

Here, $p=0.76$ , $n=25$ and $x>20$

  $q=1-p=1-0.76=0.24$

  $P(\text{ more than 20})=P(21)+P(22)+P(23)+P(24)+P(25)$

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(20)=C{}_{20}{(0.76)}^{20}{(0.24)}^{25-20}$

  $P(20)=\left(53130\right){(0.76)}^{20}{(0.24)}^{25-20}$

  $P(20)=0.17485$

  $P(21)=C{}_{21}{(0.76)}^{21}{(0.24)}^{25-21}$

  $P(21)=0.1318$

  $P(22)=C{}_{22}{(0.76)}^{22}{(0.24)}^{25-22}$

  $P(22)=0.0759$

  $P(23)=C{}_{23}{(0.76)}^{23}{(0.24)}^{25-23}$

  $P(23)=0.03135$

  $P(24)=C{}_{24}{(0.76)}^{24}{(0.24)}^{25-24}$

  $P(24)=0.00827$

  $P(25)=C{}_{25}{(0.76)}^{25}{(0.24)}^{25-25}$

  $P(25)=0.00104$

  $\begin{array}{l}P(\text{ more than 20})=P(21)+P(22)+P(23)+P(24)+P(25)\\ =0.1318+0.0759+0.03135+0.00827+0.00104\\ =0.24836\end{array}$

Conclusion:

Hence, the probability of using Google more than 20is $0.24836$

To determine

To explain: whether it is unusual if fewer than 12 used Google

Answer to Problem 31E

Yes, the probability would be unusual If fewer than 12 of them Google.

Explanation of Solution

Given:

Users of Google = 76%

Sample study = 25

Calculation:

Here, $p=0.76$ , $n=25$ and $x<12$

  $\begin{array}{l}P(\text{fewer than 12})=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+\\ P(10)+P(11)+P(12)\end{array}$

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(0)=C{}_0{(0.76)}^0{(0.24)}^{25-0}$

  $P(0)=0.32009\times {10}^{-15}$

  $P(1)=C{}_1{(0.76)}^1{(0.24)}^{25-1}$

  $P(1)=2.534\times {10}^{-14}$

  $P(2)=C{}_2{(0.76)}^2{(0.24)}^{25-2}$

  $P(2)=9.629\times {10}^{-13}$

  $P(3)=C{}_3{(0.76)}^3{(0.24)}^{25-3}$

  $P(3)=2.3378\times {10}^{-11}$

  $P(4)=C{}_4{(0.76)}^4{(0.24)}^{25-4}$

  $P(4)=4.07\times {10}^{-10}$

  $P(5)=C{}_5{(0.76)}^5{(0.24)}^{25-5}$

  $P(5)=5.415\times {10}^{-9}$

  $P(6)=C{}_6{(0.76)}^6{(0.24)}^{25-6}$

  $P(6)=5.716\times {10}^{-8}$

  $P(7)=C{}_7{(0.76)}^7{(0.24)}^{25-7}$

  $P(7)=4.913\times {10}^{-7}$

  $P(8)=C{}_8{(0.76)}^8{(0.24)}^{25-8}$

  $P(8)=3.5\times {10}^{-6}$

  $P(9)=C{}_9{(0.76)}^9{(0.24)}^{25-9}$

  $P(9)=0.0000209$

  $P(10)=C{}_{10}{(0.76)}^{10}{(0.24)}^{25-10}$

  $P(10)=0.00010$

  $P(11)=C{}_{11}{(0.76)}^{11}{(0.24)}^{25-11}$

  $P(11)=0.000458$

  $P(12)=C{}_{12}{(0.76)}^{12}{(0.24)}^{25-12}$

  $P(12)=0.00169$

  $\begin{array}{l}P(\text{fewer than 12})=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+\\ P(10)+P(11)+P(12)\\ =0.32009\times {10}^{-15}+2.534\times {10}^{-14}+9.629\times {10}^{-13}+2.3378\times {10}^{-11}+4.07\times {10}^{-10}+5.415\times {10}^{-9}+5.716\times {10}^{-8}\\ 4.913\times {10}^{-7}+3.5\times {10}^{-6}+0.0000209+0.00010+0.000458+0.00169\\ =0.00228\\ \end{array}$

Therefore, the probability that fewer than 12 of them Google is $0.00228$

Yes, the probability would be unusual If fewer than 12 of them Google.

Because, the probability that fewer than 12 of them Google is ( $0.00228$ ) small.

Conclusion:

Yes, the probability would be unusual If fewer than 12 of them Google.