Chapter 6.4, Problem 5E

(a)

To determine

The probability for $n=30,p=0.5\text{for }X=18$.

Answer to Problem 5E

The probability for $n=30,p=0.5\text{for }X=18$ is nearly 0.

Explanation of Solution

Given info: 

The value of binomial random variable X is 18 with $n=30,p=0.5$.

Calculations:

Binomial Distribution:

A random variable X is said to follow Binomial distribution if X has only two outcomes “success” and “failure” and the mass function of X is,

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

Where n is the number of trials and p is the probability of success, $q=1-p$.

The mean is $\mu =np$ and the standard deviation is $\sigma =\sqrt{(npq)}$.

Normal distribution:

A random variable x is said to follow a Normal distribution, if the probability distribution is

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-u}{\sigma }\right)}^2}$

Where,

$\mu =\text{Mean of normal variable }X$

$\sigma =\text{Standard deviation of normal variable }X$.

$\pi \text{ = 3}\text{.1415}\mathrm{...}$.

$e=\text{ 2}\text{.71828}$

Standard Normal Distribution:

A Random variable z is said to a standard normal variable if z follows normal distribution with mean $\mu =0$ and variance ${\sigma }^2=1$.

The z-score value is defined as $z=\frac{x-\mu }{\sigma }$.

The mean of binomial distribution is $\mu =np=30\left(0.5\right)=15$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(30)\left(0.5\right)\left(1-0.5\right)}=\sqrt{7.5}=2.74$.

The formula for finding the z score is,

Substitute 18 for x, 15 for $\mu$, and 2.74 for $\sigma$

$\begin{array}{c}z=\frac{18-15}{2.74}\\ =\frac{3}{2.74}\\ =1.09\end{array}$

The probability that 65 to 85 (inclusive) of them live in homes that they own is,

$\begin{array}{c}P\left(X=1.09\right)=P\left(1.09-0.5<X<1.09+0.5\right)\left(\text{Using continuity correction}\right)\\ =P\left(\frac{0.59-15}{2.74}<\frac{X-\mu }{\sigma }<\frac{1.59-15}{2.74}\right)\\ =P\left(-5.26<Z<-4.89\right)\end{array}$

The z score is obtained as follows:

For $P\left(-5.26<Z<-4.89\right)$

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 0.0 and Standard deviation as 1.0.
• Click the Shaded Area tab.
• Choose X value and Middle for the region of the curve to shade.
• Enter the X value1 as -5.26.
• Enter the X value2 as -4.89.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(-5.26<Z<-4.89\right)$ value is 0.0000004 which is nearly zero.

Thus, the probability for $n=30,p=0.5\text{for }X=18$ is nearly 0.

(b)

To determine

The probability for $n=50,p=0.8\text{for }X=44$.

Answer to Problem 5E

The probability for $n=50,p=0.8\text{for }X=44$ is 0.145.

Explanation of Solution

Given info:

The value of binomial random variable X is 44 with $n=50,p=0.8$.

Calculations:

The mean of binomial distribution is $\mu =np=50\left(0.8\right)=40$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(50)\left(0.8\right)\left(1-0.8\right)}=\sqrt{8}=2.82$.

The formula for finding the z score is,

Substitute 44 for x, 40 for $\mu$, and 2.82 for $\sigma$

$\begin{array}{c}z=\frac{44-40}{2.82}\\ =\frac{4}{2.82}\\ =1.42\end{array}$

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Calc > Probability Distributions > Normal Distribution.
• Choose Probability.
• In Mean, enter 0.
• In Standard deviation, enter 1.0.
• In Input constant, enter 1.42.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(Z=1.42\right)=0.145$

Thus, the probability for $n=50,p=0.8\text{for }X=44$ is 0.145.

(c)

To determine

The probability for $n=100,p=0.1\text{for }X=12$.

Answer to Problem 5E

The probability for $n=100,p=0.1\text{for }X=12$ is 0.318.

Explanation of Solution

Given info: 

The value of binomial random variable X is 12 with $n=100,p=0.1$.

Calculations:

The mean of binomial distribution is $\mu =np=100\left(0.1\right)=10$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(100)\left(0.1\right)\left(1-0.1\right)}=\sqrt{9}=3$.

The formula for finding the z score is,

Substitute 12 for x, 10 for $\mu$, and 3 for $\sigma$

$\begin{array}{c}z=\frac{12-10}{3}\\ =\frac{2}{3}\\ =0.67\end{array}$

Software procedure:

Step-by-step procedure to obtain the probability using the MINITAB software:

• Choose Calc > Probability Distributions > Normal Distribution.
• Choose Probability.
• In Mean, enter 0.
• In Standard deviation, enter 1.0.
• In Input constant, enter 0.67.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(Z=0.67\right)=0.318$

Thus, the probability for $n=100,p=0.1\text{for }X=12$ is 0.318.