Chapter 7.5, Problem 22E

Genetics: Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds. One hundred such offspring are produced.

1. Approximate the probability that more than 30 have green seeds.
2. Approximate the probability that 80 or fewer have yellow seeds.
3. Approximate the probability that the number with green seeds is between 30 and 35, inclusive.

To determine

The approximated probability that more than $30$ have green seeds.

Answer to Problem 22E

The approximated probability that more than $30$ have green seeds would be $0.0004$ .

Explanation of Solution

Given Information:

According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability $0.75$ of having yellow seeds and probability $0.25$ of having green seeds. One hundred such offspring are produced.

Formula used:

  $\begin{array}{l}{\mu }_x=np\\ {\sigma }_x=\sqrt{np\left(1-p\right)}\\ z=\frac{\mu -p}{\sigma }\end{array}$

Calculation:

The sample size issize is $n=100$ and population proportion is $p=25\%=0.25$ .

Since, $np=\left(100\right)\left(0.25\right)=25\ge 10$

And $n\left(1-p\right)=\left(100\right)\left(1-0.25\right)=75\ge 10$ .

This means normal approximation can be used.

  $\begin{array}{c}\text{Now, mean }{\mu }_x=np\\ =\left(100\right)\left(0.25\right)\\ =25\end{array}$

The standard deviation is,

  $\begin{array}{c}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{100\left(0.25\right)\left(1-0.25\right)}\\ =4.33\end{array}$

Since the probability is $P\left(X>30\right)$ , need to calculate the area to the left of

  $P\left(X\ge 30-0.5\right)=P\left(X\ge 39.5\right)$

Calculating the $z$ score for $39.5$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ z=\frac{39.5-25}{4.33}\\ =3.35\end{array}$

The required probability is,

  $\begin{array}{c}P\left(X\ge 39.5\right)=1-P\left(X\le 39.5\right)\\ =1-P\left(z\le 3.35\right)\\ =1-0.9996\\ =0.0004\end{array}$

Hence, the approximated probability that more than $30$ have green seeds would be $0.0004$ .

To determine

The approximated probability that $80$ or fewer have yellow seeds

Answer to Problem 22E

The approximated probability that $80$ or fewer have yellow seeds would be $0.8980$ .

Explanation of Solution

Given Information:

The sample size is $n=100$ and population proportion for yellow seeds is $p=0.75$ .

Calculation:

The sample size issize is $n=100$ and population proportion is $p=75\%=0.75$ .

Since, $np=\left(100\right)\left(0.75\right)=75\ge 10$

And $n\left(1-p\right)=\left(100\right)\left(1-0.75\right)=25\ge 10$ .

This means normal approximation can be used.

  $\begin{array}{c}\text{Now, mean }{\mu }_x=np\\ =\left(100\right)\left(0.75\right)\\ =75\end{array}$

The standard deviation is,

  $\begin{array}{c}{\sigma }_x=\sqrt{np\left(1-p\right)}\\ =\sqrt{100\left(0.75\right)\left(1-0.75\right)}\\ =4.33\end{array}$

Since the probability is $P\left(X\le 80\right)$ , need to calculate compute the area to the left of

  $P\left(X\le 80+0.5\right)=P\left(X\le 80.5\right)$

Calculating the $z$ score for $80.5$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ z=\frac{80.5-75}{4.33}\\ =1.27\end{array}$

The required probability is,

  $\begin{array}{l}P\left(X\le 80.5\right)\\ =P\left(z\le 1.27\right)\\ =0.8980\end{array}$

Hence, the approximated probability that $80$ or fewer have yellow seeds would be $0.8980$ .

To determine

The approximated probability that the number with green seeds is between $30\text{ and 35}$ inclusive.

Answer to Problem 22E

The approximated probability that the number with green seeds is between $30\text{ and 35}$ inclusive would be $0.1414$ .

Explanation of Solution

Given Information:

The sample size is $n=100$ and population proportion for yellow seeds is $p=0.75$ .

Calculation:

Since, the probability is $P\left(30\le X\le 35\right)$ , need to calculate the area between,

  $P\left(30-0.5\le X\le 35+0.5\right)=P\left(29.5\le X\le 35.5\right)$

Calculating the $z$ score for $29.5$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ z=\frac{29.5-25}{4.33}\\ =1.04\end{array}$

Calculating the $z$ score for $35.5$ as below,

  $\begin{array}{c}z=\frac{\widehat{p}-{\mu }_p}{{\sigma }_p}\\ z=\frac{35.5-25}{4.33}\\ =2.42\end{array}$

The required probability is

  $\begin{array}{c}P\left(29.5\le X\le 35.5\right)\\ =P\left(X\le 35.5\right)-P\left(X\le 29.5\right)\\ =P\left(z\le 2.42\right)-P\left(z\le 1.04\right)\\ =0.9922-0.8508\\ =0.1414\end{array}$

Hence, the approximated probability that the number with green seeds is between $30\text{ and 35}$ inclusive would be $0.1414$ .