Chapter 6.4, Problem 6E

(a)

To determine

The probability for $n=10,p=0.5\text{for }X\ge 7$.

Answer to Problem 6E

The probability for $n=10,p=0.5\text{for }X\ge 7$ is 0.1038.

Explanation of Solution

Given info:

The value of binomial random variable X is greater than 7 with $n=10,p=0.5$.

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 $\mu =np=10\left(0.5\right)=5$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(10)\left(0.5\right)\left(1-0.5\right)}=\sqrt{2.5}=1.58$.

The formula for finding the z score is,

Substitute 7 for x, 10 for $\mu$, and 1.58 for $\sigma$

$\begin{array}{c}z=\frac{7-5}{1.58}\\ =\frac{2}{1.58}\\ =1.26\end{array}$

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 Right Tail for the region of the curve to shade.
• Enter the X value as 1.26.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(Z\ge 1.26\right)=0.1038$

Thus, the probability for $n=30,p=0.5\text{for }X\ge 7$ is 0.1038.

 (b)

To determine

The probability for $n=20,p=0.7\text{for }X\le 12$.

Answer to Problem 6E

The probability for $n=20,p=0.7\text{for }X\le 12$ is 0.1660.

Explanation of Solution

Given info: 

The value of binomial random variable X is less than 12 with $n=20,p=0.7$.

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 $\mu =np=20\left(0.7\right)=14$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(20)\left(0.7\right)\left(1-0.3\right)}=\sqrt{4.2}=2.05$.

The formula for finding the z score is,

Substitute 12 for x, 14 for $\mu$, and 2.05 for $\sigma$

$\begin{array}{c}z=\frac{12-14}{2.05}\\ =-\frac{2}{2.05}\\ =-0.97\end{array}$

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 Left Tail for the region of the curve to shade.
• Enter the X value as -0.97.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(Z\le -0.97\right)=0.1660$

Thus, the probability for $n=20,p=0.7\text{for }X\le 12$ is 0.1660.

(c)

To determine

The probability for $n=50,p=0.6\text{for }X\le 40$.

Answer to Problem 6E

The probability for $n=50,p=0.6\text{for }X\le 40$ is 1.

Explanation of Solution

Given info:

The value of binomial random variable X is less than 40 with $n=50,p=0.6$.

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 $\mu =np=50\left(0.6\right)=30$.

The standard deviation of binomial distribution is $\sigma =\sqrt{(50)\left(0.6\right)\left(1-0.6\right)}=\sqrt{12}=3.46$.

The formula for finding the z score is,

Substitute 40 for x, 30 for $\mu$, and 1.09 for $\sigma$

$\begin{array}{c}z=\frac{40-30}{3.46}\\ =\frac{10}{3.46}\\ =\mathrm{2.89.}\end{array}$

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 Left Tail for the region of the curve to shade.
• Enter the X value as 2.89.
• Click OK.

Output using the MINITAB software is given below:

From MINITAB output, $P\left(Z\le 9.13\right)=1$

Thus, the probability for $n=50,p=0.6\text{for }X\le 40$ is 1.