Chapter 5, Problem 2DH

a

To determine

Find the probability that out of 10 adults selected at random, at least half of them who would agree that humanity should explore the planet.

Answer to Problem 2DH

The probability that out of 10 adults selected at random, at least half who would agree that humanity should explore the planet is 0.906.

Explanation of Solution

The number of adult who agrees that humanity should explore the planet follows a binomial distribution with $n=10\text{ and }p\text{=0}\text{.65}$.

Step-by-step procedure to obtain the probability using the binomial probability table:

• Locate $n=10\text{ and }p\text{=0}\text{.65}$ in the Table 3 of binomial probability distribution in Appendix II.
• Obtain the value in the corresponding row $\left(r=5,6,7,8,9,10\right)$ below 0.65.

The required probability values from the binomial probability distribution table are as follows:

$\begin{array}{l}P\left(X=5\right)=0.154,P\left(X=6\right)=0.238,P\left(X=7\right)=0.252,\\ P\left(X=8\right)=0.176,P\left(X=9\right)=0.072,P\left(X=10\right)=0.014\end{array}$

The probability that half of the selected adults who agree that humanity should explore the planet is as follows:

$\begin{array}{c}P\left(X\ge 5\right)=P\left(X=5\right)+P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right)+P\left(X=9\right)+P\left(X=10\right)\\ =0.154+0.238+0.252+0.176+0.072+0.014\\ =0.906\end{array}$

Thus, the probability that out of 10 adults selected at random, at least half of them who would agree that humanity should explore the planet is 0.906.

b

To determine

Compute the probability that out of 10 adults selected at random, at least 9 of them who would agree that space exploration does have an impact on daily life.

Answer to Problem 2DH

The probability that out of 10 adults selected at random, at least 9 of them who would agree that space exploration does have an impact on daily life is 0.0309.

Explanation of Solution

The number of adults who agree that space exploration does have an impact on daily life follows a binomial distribution with $n=10\text{ and }p\text{=0}\text{.57}$.

The probability that at least 9 of them who would agree that space exploration does have an impact on daily life is as follows:

$\begin{array}{c}P\left(X\ge 9\right)=P\left(X=9\right)+P\left(X=10\right)\\ =C_{10,9}\times {0.57}^9\times 0.43+C_{10,10}\times {0.57}^{10}\times {0.43}^0\\ =\frac{10!}{9!\times 1!}\times {0.57}^9\times 0.43+\frac{10!}{10!\times 0!}\times {0.57}^{10}\times {0.43}^0\\ =0.0273+0.0036\\ =0.0309\end{array}$

Thus, the probability that out of 10 adults selected at random, at least 9 of them who would agree that space exploration has an impact on daily life is 0.0309.

c

To determine

Compute the expected value and standard deviation of adults who would travel in the space, given a chance.

Answer to Problem 2DH

The expected value is 15.4.

The standard deviation is 2.9367.

Explanation of Solution

The number of adults who would travel in the space, given a chance follows a binomial distribution with $n=35\text{ and }p\text{=0}\text{.44}$.

The expected value and standard deviation are calculated as follows:

$\begin{array}{c}E\left(X\right)=np\\ =35\times 0.44\\ =15.4\\ \sigma =\sqrt{npq}\\ =\sqrt{35\times 0.44\times \left(1-0.44\right)}\\ =2.9367\end{array}$

Thus, the expected value and standard deviation are 15.4 and 2.9367, respectively.

d

To determine

Find the probability that the first adult selected would agree with the statement that space will be colonized in a person’s life time.

Answer to Problem 2DH

The probability that the first adult selected would agree with the statement that space will be colonized in a person’s life time is 0.18.

Explanation of Solution

The required probability is calculated as follows:

$\begin{array}{c}P\left(\text{first adult agree}\right)=p{q}^{x-1}\\ =0.18\times {\left(1-0.18\right)}^{1-1}\\ =0.18\end{array}$

Thus, the probability that the first adult selected would agree with the statement that space will be colonized in a person’s life time is 0.18.