Chapter 6.4, Problem 22E

a.

To determine

Find the probability that more than 30 have green seed by using Approximation.

Answer to Problem 22E

The probability that more than 30 have green seed by using Approximation is 0.1020.

Explanation of Solution

Calculation:

The given information is that “when two heterozygous plants are crossed then the plants offspring will have probability 0.75 of having yellow seeds and 0.25 of having green seeds”. The number of offspring produced is 100.

Normal approximation to the Binomial:

If $np\ge 10$ and $n\left(1-p\right)\ge 10$ where n is the number of trials and p is the success probability then the binomial random variable X follows approximately normal distribution with mean ${\mu }_X=np$ and standard deviation ${\sigma }_X=\sqrt{np\left(1-p\right)}$.

Let X represents the number of offspring produced by plants. Therefore, the probability that more than 30 have green seed represents $P\left(X>30\right)$.

From the given information, it can be observed that the sample size n is 100 and the success probability of yellow seeds is 0.75 and success probability of green seeds is 0.25.

Requirement check for green seeds:

Condition 1: $np\ge 10$

Condition 2: $n\left(1-p\right)\ge 10$

Condition 1: $np\ge 10$

Substitute 100 for n and 0.25 for p in np,

$\begin{array}{c}np=\left(100\right)\left(0.25\right)\\ =25\end{array}$

Thus, the requirement $np\left(=25\right)\ge 10$ is satisfied.

Condition 2: $n\left(1-p\right)\ge 10$

Substitute 100 for n and 0.25 for p in $n\left(1-p\right)$,

$\begin{array}{c}n\left(1-p\right)=\left(100\right)\left(1-0.25\right)\\ =100\left(0.75\right)\\ =75\end{array}$

Thus, the requirement of $n\left(1-p\right)\left(=75\right)\ge 10$ is satisfied.

Here, both the requirements are satisfied. Thus, it is appropriate to use the normal approximation.

For mean ${\mu }_X$:

Substitute 100 for n and 0.25 for p in the formula of ${\mu }_X$,

$\begin{array}{c}{\mu }_X=\left(100\right)\left(0.25\right)\\ =25\end{array}$

Thus, the value of ${\mu }_X$ is 25.

For standard deviation ${\sigma }_X$:

Substitute 100 for n and 0.25 in the formula of ${\sigma }_X$,

$\begin{array}{c}{\sigma }_X=\sqrt{100\left(0.25\right)\left(1-0.25\right)}\\ =\sqrt{100\left(0.25\right)\left(0.75\right)}\\ =\sqrt{18.75}\\ \approx 4.33013\end{array}$

Thus, the value of ${\sigma }_X$ is 4.33013.

Continuity correction:

The continuity correction is the adjustment while approximating a discrete distribution with a continuous distribution. The areas under the normal curve to use when the continuity correction applied to the probabilities are as follows:

• If $P\left(a\le X\le b\right)$ then find area between $a-0.5$ and $b+0.5$.
• $P\left(X\le b\right)$ then find area to the left of $b+0.5$.
• $P\left(X\ge a\right)$ then find area to the right of $a-0.5$.
• $P\left(X=a\right)$ then find area between $a-0.5$ and $a+0.5$.

Here, probability $P\left(X>30\right)$ is equal to $P\left(X\ge 31\right)$.

Also, it can be observed that the probability $P\left(X\ge 31\right)$ represents $P\left(X\ge a\right)$ by using continuity correction. Therefore, probability $P\left(X\ge 31\right)$ is the area to the right of 30.5 $\left(=31-0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 25 and Standard deviation as 4.33013.
• Click the Shaded Area tab.
• Choose X value and Right Tail for the region of the curve to shade.
• Enter the X value as 30.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability that more than 30 have green seed by using Approximation is 0.1020.

b.

To determine

Find the probability 80 or fewer have yellow seeds by using Approximation.

Answer to Problem 22E

The probability 80 or fewer have yellow seeds by using Approximation is 0.8980.

Explanation of Solution

Calculation:

The probability 80 or fewer have yellow seeds represents $P\left(X\le 80\right)$.

Requirement check for yellow seeds:

Condition 1: $np\ge 10$

Condition 2: $n\left(1-p\right)\ge 10$

Condition 1: $np\ge 10$

Substitute 100 for n and 0.75 for p in np,

$\begin{array}{c}np=\left(100\right)\left(0.75\right)\\ =75\end{array}$

Thus, the requirement $np\left(=75\right)\ge 10$ is satisfied.

Condition 2: $n\left(1-p\right)\ge 10$

Substitute 100 for n and 0.75 for p in $n\left(1-p\right)$,

$\begin{array}{c}n\left(1-p\right)=\left(100\right)\left(1-0.75\right)\\ =100\left(0.25\right)\\ =25\end{array}$

Thus, the requirement of $n\left(1-p\right)\left(=25\right)\ge 10$ is satisfied.

Thus, both the requirements are satisfied. Thus, it is appropriate to use the normal approximation.

For mean ${\mu }_X$:

Substitute 100 for n and 0.75 for p in the formula of ${\mu }_X$,

$\begin{array}{c}{\mu }_X=\left(100\right)\left(0.75\right)\\ =75\end{array}$

Thus, the value of ${\mu }_X$ is 75.

For standard deviation ${\sigma }_X$:

Substitute 100 for n and 0.75 in the formula of ${\sigma }_X$,

$\begin{array}{c}{\sigma }_X=\sqrt{100\left(0.75\right)\left(1-0.75\right)}\\ =\sqrt{100\left(0.75\right)\left(0.25\right)}\\ =\sqrt{18.75}\\ \approx 4.33013\end{array}$

Thus, the value of ${\sigma }_X$ is 4.33013.

Here, the probability $P\left(X\le 80\right)$ represents $P\left(X\le b\right)$. Therefore, probability $P\left(X\le 80\right)$ is the area to the left of 80.5 $\left(=80+0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 75 and Standard deviation as 4.33013.
• Click the Shaded Area tab.
• Choose X value and Left Tail for the region of the curve to shade.
• Enter the X value as 80.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability 80 or fewer have yellow seeds by using Approximation is 0.8980.

c.

To determine

Find the probability that the number with green seeds is between 30 and 35, inclusive by using approximation.

Answer to Problem 22E

The probability that the number with green seeds is between 30 and 35, inclusive by using approximation is 0.1417.

Explanation of Solution

Calculation:

The probability that the number with green seeds is between 30 and 35, inclusive represents $P\left(30\le X\le 35\right)$.

Also, it can be observed that the probability $P\left(30\le X\le 35\right)$ represents $P\left(a\le X\le b\right)$ by using continuity correction. Therefore, probability $P\left(30\le X\le 35\right)$ is the area between 29.5 $\left(=30-0.5\right)$ and 35.5 $\left(=35+0.5\right)$.

Software Procedure:

Step by step procedure to find the probability by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 25 and Standard deviation as 4.33013.
• Click the Shaded Area tab.
• Choose X value and Middle for the region of the curve to shade.
• Enter the X value 1 as 29.5 and X value 2 as 35.5.
• Click OK.

Output using MINITAB software is as follows:

From the output, it can be observed that the probability that the number with green seeds is between 30 and 35, inclusive by using approximation is 0.1417.