Chapter 10, Problem 1CA

(a)

To determine

To graph:

Ahistogram for the binomial distribution of survey.

Explanation of Solution

Given information:

  $\begin{array}{l}n=14\\ p=0.63\\ k=0,1,2,\mathrm{....},14\end{array}$

Graph:

  

Histogram ofprobability distribution

Interpretation:

For a binomial experiment, the probability of $k$ successes in $n$ trials can be given as:

  $P(ksuccesses)=Cp{}^k{(1-p)}^{n-k}$

Where,

  $p=$ Probability of the success.

From the given information,

Substitute the values of $n,p,k$ in the formula stated above.

The required probability is as follows:

For $k=0$ ,

  $\begin{array}{l}$ P(k=0)=C(0.63){}^0{(1-0.63)}^{14-0}$$ P(k=0)=C(0.63){}^0{(0.37)}^{14}$P(k=0)=1\times 1\times 9.012\times {10}^{-7}\\ P(k=0)=9.01\times {10}^{-7}\\ P(k=0)\simeq 0.0000\end{array}$

For $k=1$ ,

  $\begin{array}{l}$ P(k=1)=C(0.63){}^1{(1-0.63)}^{14-1}$$ P(k=1)=C\times 0.63\times {(0.37)}^{13}$P(k=1)=14\times 0.63\times 2.435\times {10}^{-6}\\ P(k=1)=14\times 1.535\times {10}^{-6}\\ P(k=1)=2.148\times {10}^{-6}\\ P(k=1)\simeq 0.0000\end{array}$

For $k=2$ ,

  $\begin{array}{l}$ P(k=2)=C(0.63){}^2{(1-0.63)}^{14-2}$$ P(k=2)=C\times 0.3969\times {(0.37)}^{12}$P(k=2)=91\times 0.3969\times 6.583\times {10}^{-6}\\ P(k=2)=91\times 2.613\times {10}^{-6}\\ P(k=2)=239.421\times {10}^{-6}\\ P(k=2)\simeq 0.00002\end{array}$

For $k=3$ ,

  $\begin{array}{l}$ P(k=3)=C(0.63){}^3{(1-0.63)}^{14-3}$$ P(k=3)=C\times 0.2501\times {(0.37)}^{11}$P(k=3)=364\times 0.2501\times 1.779\times {10}^{-5}\\ P(k=3)=364\times 4.449\times {10}^{-6}\\ P(k=3)=1.6195\times {10}^{-3}\\ P(k=3)\simeq 0.0016\end{array}$

Similarly, find the values for $k=4,5,6,\mathrm{.....},14$

The values are:

  $\begin{array}{l}P(k=4)=0.0076\\ P(k=5)=0.0258\\ P(k=6)=0.0659\\ P(k=7)=0.1283\\ P(k=8)=0.1912\\ P(k=9)=0.2170\\ P(k=10)=0.1848\\ P(k=11)=0.1144\\ P(k=12)=0.0487\\ P(k=13)=0.0128\\ P(k=14)=0.0016\end{array}$

From the interpretation, histogram showing the random variable probability distribution is shown above.

(b)

To determine

To calculate:

The most likely number of Americans that are considering themselves as sports fans.

Answer to Problem 1CA

The most likely number of Americans that are considering themselves as sports fans are $9$ out of $14$ .

Explanation of Solution

Given information:

Random selection for the survey $=14$ .

Calculation:

For the survey results, the probability distribution shows that the probability is greatest for $9$ , which is more than 0.2.

i.e. from subpart (a),

  $P(k=9)=0.2170$

Thus, the most likely number of Americans that are considering themselves as sports fans are $9$ .

(c)

To determine

To calculate:

The probability of atleast $7$ Americans out of $14$ considering themselves as sports fan.

Answer to Problem 1CA

The probability of at least $7$ Americans considering themselves as sports fan out of $14$ is $0.8988$ .

Explanation of Solution

Given information:

From subpart (a), the values of probabilities are:

  $\begin{array}{l}P(k=7)=0.1283\\ P(k=8)=0.1912\\ P(k=9)=0.2170\\ P(k=10)=0.1848\\ P(k=11)=0.1144\\ P(k=12)=0.0487\\ P(k=13)=0.0128\\ P(k=14)=0.0016\end{array}$

Calculation:

The probability of at least 7 Americans out of 14 can be given as:

  $P(k\ge 7)=P(k=7)+P(k=8)+P(k=9)+P(k=10)+P(k=11)+P(k=12)+P(k=13)+P(k=14)$

Substitute the respective values in above equation.

  $\begin{array}{l}P(k\ge 7)=0.1283+0.1912+0.2170+0.1848+0.1144+0.0487+0.0128+0.0016\\ P(k\ge 7)=0.8988\end{array}$

Thus, the probability of at least $7$ Americans considering themselves as sports fan out of $14$ is $0.8988$ .