Chapter 5, Problem 19P

a.

To determine

Find the value of $f\left(0\right)$.

Answer to Problem 19P

The value of $f\left(0\right)$is 0.3487.

Explanation of Solution

Calculation:

The binomial experiment with$n=10\text{ and }p=0.10$.

The probability of obtaining x successes in n independent trails of a binomial experiment is,

$f\left(x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,1,2,\mathrm{...},n$

Where, p is the probability of success.

Substitute $n=10\text{ and }p=0.10\text{ and }x=0$

$\begin{array}{c}f\left(0\right)=\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.10\right)}^0{\left(1-0.10\right)}^{10-0}\\ =\frac{10!}{0!\left(10-0\right)!}{\left(0.10\right)}^0{\left(0.9\right)}^{10}\\ =0.3487\end{array}$

Thus, the probability of zero success is 0.3487.

b.

To determine

Find the value of $f\left(2\right)$.

Answer to Problem 19P

The value of $f\left(2\right)$is 0.1937.

Explanation of Solution

Calculation:

Consider $n=10\text{ and }p=0.10\text{ and }x=2$

The value of $f\left(2\right)$ is

$\begin{array}{c}f\left(2\right)=\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.10\right)}^2{\left(1-0.10\right)}^{10-2}\\ =\frac{10!}{2!\left(10-2\right)!}{\left(0.10\right)}^2{\left(0.9\right)}^8\\ =0.1937\end{array}$

Thus, the probability of two successes is 0.1937.

c.

To determine

Find the probability of at most two successes.

Answer to Problem 19P

The probability of at most two successes is 0.9298.

Explanation of Solution

Calculation:

Here,$P\left(x\le 2\right)$is the probability of at most two successes. This probability is the sum of the probabilities at the points$x=0,x=1\text{ and }x=2$. That is,

$P\left(x\le 2\right)=f\left(0\right)+f\left(1\right)+f(2)$

The probability $f\left(1\right)$ is obtained by substituting the values $n=10\text{ and }p=0.10\text{ and }x=1$

in the binomial probability formula. That is,

$\begin{array}{c}f\left(1\right)=\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.10\right)}^1{\left(1-0.10\right)}^{10-1}\\ =\frac{10!}{1!\left(10-1\right)!}{\left(0.10\right)}^1{\left(0.9\right)}^9\\ =0.3874\end{array}$

Substitute the other probability values from part (a) and (b),

$\begin{array}{c}P\left(x\le 2\right)=f\left(0\right)+f\left(1\right)+f(2)\\ =0.3487+0.3874+0.1937\\ =0.9298\end{array}$

Thus, the probability of at most two successes is 0.9298.

d.

To determine

Find the probability of at least one success.

Answer to Problem 19P

The probability of at least one success is 0.6513.

Explanation of Solution

Calculation:

Here, $P\left(x\ge 1\right)$is the probability of at least one success can be obtained as the compliment the probability$P\left(x<1\right)$. That is,

$\begin{array}{c}P\left(x\ge 1\right)=1-P\left(x<1\right)\\ =1-\left[f\left(0\right)\right]\end{array}$

Substitute the value from part (a),

$\begin{array}{c}P\left(x\ge 1\right)=1-\left[f\left(0\right)\right]\\ =1-0.3487\\ =0.6513\end{array}$

Thus, the probability of at least one success is 0.6513.

e.

To determine

Find the expected value.

Answer to Problem 19P

The expected value of the binomial random variable is 1.

Explanation of Solution

Calculation:

The expected value of a binomial random variable is given by,

$\begin{array}{c}E\left(x\right)=\mu \\ =np\end{array}$

Substitute the values $n=10\text{ and }p=0.1$

$\begin{array}{c}E\left(x\right)=10\times 0.1\\ =1\end{array}$

Thus, the expected value of the binomial random variable is 1.

f.

To determine

Find the variance and standard deviation.

Answer to Problem 19P

The variance of the binomial random variable is 0.9.

The standard deviation of the binomial random variable is 0.95.

Explanation of Solution

Calculation:

The variance of a binomial random variable is given by,

${\sigma }^2=np\left(1-p\right)$

Substituting the values $n=10\text{ and }p=0.1$

$\begin{array}{c}{\sigma }^2=np\left(1-p\right)\\ =10\times 0.1\left(1-0.1\right)\\ =0.9\end{array}$

The variance of the binomial random variable is 0.9.

The standard deviation of the random variable x is obtained by taking the square root of variance.

Thus, the standard deviation of the binomial random variable is given by,

$\begin{array}{c}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{0.9}\\ =0.95\end{array}$

The standard deviation of the binomial random variable is 0.95.