Chapter 6.2, Problem 37E

College bound: The Statistical Abstract of the United States reported that 66% of students who graduated from high school in 2012 enroled in college. Thirty high school graduates are sampled.

What is the probability that exactly 15 of them enroll in college?

What is the probability that more than 15 enroll in college?

What is the probability that fewer than 12 enroll in college?

Would it be unusual if more than 25 of them enroll in college?

What is the mean number who enroll in college in a sample of 30 high school graduates?

What is the standard deviation of the number who enrol in college in a sample of 30 high school graduates?

To determine

To find: the probability of exactly 18 students who enrolled in college

Answer to Problem 37E

The probability of exactly 18 students who enrolled in college is $0.116552$

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

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, $p=0.66$ , $n=30$ and $x=18$

  $q=1-p=1-0.66=0.34$

Probability of binomial distribution is,

  $\begin{array}{l}P(x)=C{}_x{p}^x{q}^{n-x}\\ $ P(18)=C{}_{18}{(0.66)}^{18}{(0.34)}^{30-18}$P(18)=0.116552\end{array}$

Conclusion:

Therefore, the probability of exactly 18 students who enrolled in college is $0.116552$

To determine

To find: the probability of more than 15 students who enrolled in college

Answer to Problem 37E

The probability of more than 15 students who enrolled in college is $0.9486$

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

Calculation:

Here, $p=0.66$ , $n=30$ and $x<15$

  $q=1-p=1-0.66=0.34$

  $\begin{array}{l}P(\text{more than 15})=P(16)+P(17)+P(18)+P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\\ P(26)+P(27)+P(28)+P(29)+P(30)\end{array}$

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(16)=C{}_{16}{(0.66)}^{16}{(0.34)}^{30-16}$

  $P(16)=0.0520044$

  $P(17)=C{}_{17}{(0.66)}^{17}{(0.34)}^{30-17}$

  $P(17)=0.0831350$

  $P(18)=C{}_{18}{(0.66)}^{18}{(0.34)}^{30-18}$

  $P(18)=0.116552$

  $P(19)=C{}_{19}{(0.66)}^{19}{(0.34)}^{30-19}$

  $P(19)=0.142894$

  $P(20)=C{}_{20}{(0.66)}^{20}{(0.34)}^{30-20}$

  $P(20)=0.152560$

  $P(21)=C{}_{21}{(0.66)}^{21}{(0.34)}^{30-21}$

  $P(21)=0.141022$

  $P(22)=C{}_{22}{(0.66)}^{22}{(0.34)}^{30-22}$

  $P(22)=0.111988$

  $P(23)=C{}_{23}{(0.66)}^{23}{(0.34)}^{30-23}$

  $P(23)=0.0756133$

  $P(24)=C{}_{24}{(0.66)}^{24}{(0.34)}^{30-24}$

  $P(24)=0.0428105$

  $P(25)=C{}_{25}{(0.66)}^{25}{(0.34)}^{30-25}$

  $P(25)=0.0199446$

  $P(26)=C{}_{26}{(0.66)}^{26}{(0.34)}^{30-26}$

  $P(26)=0.0074454$

  $P(27)=C{}_{27}{(0.66)}^{27}{(0.34)}^{30-27}$

  $P(27)=0.0021412$

  $P(28)=C{}_{28}{(0.66)}^{28}{(0.34)}^{30-28}$

  $P(28)=0.0004453$

  $P(29)=C{}_{29}{(0.66)}^{29}{(0.34)}^{30-29}$

  $P(29)=0.0000596$

  $P(30)=C{}_{30}{(0.66)}^{30}{(0.34)}^{30-30}$

  $P(30)=0.0000039$

  $\begin{array}{l}P(\text{more than 15})=P(16)+P(17)+P(18)+P(19)+P(20)+P(21)+P(22)+P(23)+P(24)+P(25)\\ P(26)+P(27)+P(28)+P(29)+P(30)\\ =0.116552+0.0520044+0.0831350+0.142894+0.152560+0.141022+0.111988+0.0756133\\ 0.0428105+0.0199446+0.0074454+0.0021412+0.0004453+0.0000596+0.0000039\\ =0.9486\\ \\ \end{array}$

Conclusion:

Therefore, the probability of more than 15 students who enrolled in college is $0.9486$

To determine

To find: the probability of fewer than 12students who enrolled in college

Answer to Problem 37E

The probability offewer than 12students who enrolled in college is $0.0009588$

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

Calculation:

Here, $p=0.66$ , $n=30$ and $x>12$

  $q=1-p=1-0.66=0.34$

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

Probability of binomial distribution is,

  $P(x)=C{}_x{p}^x{q}^{n-x}$

  $P(0)=C{}_0{(0.66)}^0{(0.34)}^{30-0}$

  $P(0)=0.00000$

  $P(1)=C{}_1{(0.66)}^1{(0.34)}^{30-1}$

  $P(1)=0.00000$

  $P(2)=C{}_2{(0.66)}^2{(0.34)}^{30-2}$

  $P(2)=0.00000$

  $P(3)=C{}_3{(0.66)}^3{(0.34)}^{30-3}$

  $P(3)=0.00000$

  $P(4)=C{}_4{(0.66)}^4{(0.34)}^{30-4}$

  $P(4)=0.00000$

  $P(5)=C{}_5{(0.66)}^5{(0.34)}^{30-5}$

  $P(5)=0.00000$

  $P(6)=C{}_6{(0.66)}^6{(0.34)}^{30-6}$

  $P(6)=0.00000$

  $P(7)=C{}_7{(0.66)}^7{(0.34)}^{30-7}$

  $P(7)=0.00000$

  $P(8)=C{}_8{(0.66)}^8{(0.34)}^{30-8}$

  $P(8)=0.00000$

  $P(9)=C{}_9{(0.66)}^9{(0.34)}^{30-9}$

  $P(9)=0.0000493$

  $P(10)=C{}_{10}{(0.66)}^{10}{(0.34)}^{30-10}$

  $P(10)=0.0002008$

  $P(11)=C{}_{11}{(0.66)}^{11}{(0.34)}^{30-11}$

  $P(11)=0.0007087$

  $\begin{array}{l}P(fewer\text{ than 12})=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)\\ P(9)+P(10)+P(11)\\ =0.00000+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000+0.00000\\ 0.00000+0.0000493+0.0002008+0.0007087\\ =0.0009588\end{array}$

Conclusion:

Therefore, the probability offewer than 12students who enrolled in college is $0.0009588$

To determine

To explain: whether it is unusual if more than 25 students who enrolled in college

Answer to Problem 37E

It is not unusual if more than 25 students who enrolled in college

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

Calculation:

Here, $p=0.66$ , $n=30$ and $x=18$

  

  $\begin{array}{l}P(\text{more than 25})=P(26)+P(27)+P(28)+P(29)+P(30)\\ =0.0074454+0.0021412+0.0004453+0.0000596+0.0000039\\ =0.0100954\\ \end{array}$

The required probability is not low.

So, it would not be unusual if more than 25 students who enrolled in college

Conclusion:

Therefore, it is not unusual if more than 25 students who enrolled in college

To determine

To find: mean number if a sample of30 students who enrolled in college

Answer to Problem 37E

The mean number if a sample of30 students who enrolled in college is $19.8$

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

Calculation:

The mean number is,

  $\begin{array}{l}{\mu }_x=np\\ =30\left(0.66\right)\\ =19.8\end{array}$

Conclusion:

Therefore, the mean number if a sample of30 students who enrolled in college is $19.8$

To determine

To find: standard deviation if a sample of30 students who enrolled in college

Answer to Problem 37E

The standard deviation if a sample of30 students who enrolled in college is $6.732$

Explanation of Solution

Given:

Number of students enrolled in college is, $p=0.66$

Samples taken, $n=30$

Calculation:

The standard deviation is,

  $\begin{array}{c}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{30\left(0.66\right)\left(1-0.66\right)}\\ =6.732\end{array}$

Conclusion:

Therefore, the standard deviation if a sample of30 students who enrolled in college is $6.732$