Chapter 4, Problem 3SE

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.

a.    Out of 10 offspring of heterozygous plants, what is the probability that exactly 3 have green seeds?

b.    Out of 10 offspring of heterozygous plants, what is the probability that more than 2 have green seeds?

c.    Out of 100 offspring of heterozygous plants, what is the probability that more than 30 have green seeds?

d.    Out of 100 offspring of heterozygous plants, what is the probability that between 30 and 35 inclusive have green seeds?

e.    Out of 100 offspring of heterozygous plants, what is the probability that fewer than 80 have yellow seeds?

To determine

Find the probability that exactly 3 have green seeds, out of 10 off springs of heterozygous plants.

Answer to Problem 3SE

The probability that exactly 3 have green seeds is 0.2503.

Explanation of Solution

Given info:

The probability of each of the off springs having yellow seeds is 0.75 and having green seeds is 0.25.

Calculation:

The random variable X is defined as the number of plants out of 10 that have green seeds. Then the random variable follows binomial with parameters $n=10,x=3\text{and }p=0.25$.

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

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

Where, p is the probability of success.

Substitute $n=10,x=3\text{and }p=0.25$

$\begin{array}{c}P\left(X=3\right)=\left(\begin{array}{l}10\\ 3\end{array}\right){\left(0.25\right)}^3{\left(1-0.25\right)}^{10-3}\\ =\frac{10!}{3!\left(10-3\right)!}{\left(0.25\right)}^3{\left(0.75\right)}^7\\ =\left(120\right)\left(0.01563\right)\left(0.13348\right)\\ =0.2503\end{array}$

Thus, the value of $P\left(X=3\right)=0.2503$.

To determine

Find the probability that more than 2 have green seeds, out of 10 off spring of heterozygous plants.

Answer to Problem 3SE

The probability that more than 2 have green seeds is 0.4744.

Explanation of Solution

Calculation:

$P\left(X>2\right)=1-P\left(X\le 2\right)$

Substitute $n=10,x=2\text{and }p=0.25$

$\begin{array}{c}P\left(X>2\right)=1-P\left(X=0\right)-P\left(X=1\right)-P\left(X=2\right)\\ =\left(\begin{array}{c}1-\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.25\right)}^0{\left(1-0.25\right)}^{10-0}-\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.25\right)}^1{\left(1-0.25\right)}^{10-1}\\ +\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.25\right)}^2{\left(1-0.25\right)}^{8-2}\end{array}\right)\\ =\left(\begin{array}{c}1-\frac{10!}{0!\left(10-0\right)!}{\left(0.25\right)}^0{\left(0.75\right)}^{10}-\frac{10!}{1!\left(10-1\right)!}{\left(0.25\right)}^1{\left(0.75\right)}^9\\ +\frac{10!}{2!\left(10-2\right)!}{\left(0.25\right)}^2{\left(0.75\right)}^8\end{array}\right)\\ =1-\left(1\right)\left(1\right)\left(0.0563\right)-\left(10\right)\left(0.25\right)\left(0.0751\right)-\left(45\right)\left(0.0625\right)\left(0.1001\right)\end{array}$

$\begin{array}{c}=1-0.0563-0.1878-0.2816\\ =0.4744\end{array}$

Thus, the probability that more than 2 have green seeds is 0.4744.

To determine

Find the probability that more than 30 have green seeds, out of 100 off spring of heterozygous plants.

Answer to Problem 3SE

The probability that more than 30 have green seeds is 0.1020.

Explanation of Solution

Calculation:

The random variable Y is defined as the number of plants out of 100 that have green seeds. Then the random variable Y follows binomial distribution with parameters $n=100\text{ and }p=0.25$.

The condition for the normal approximation to the binomial probability distribution is $np>10\text{ and }n\left(1-p\right)>10$.

That is,

$\begin{array}{c}np=100\left(0.25\right)\\ =25\\ >10\end{array}$

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

Thus, the normal distribution can be used to approximate the binomial distribution.

Mean:

${\mu }_Y=np$

Substitute 100 for n and 0.25 for p.

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

Standard deviation:

${\sigma }_Y=\sqrt{np(1-p)}$

Substitute n as 100and p as 0.25.

$\begin{array}{c}{\sigma }_Y=\sqrt{\left(100\right)\left(0.25\right)\left(1-0.25\right)}\\ =\sqrt{18.75}\\ =4.3301\end{array}$

The approximate binomial probability using the normal distribution is,

$P\left(Y>a\right)\approx P\left(Y\ge a+0.5\right)$

By using continuity correction, the value 0.5 is added from the value of 30.

Substitute 30 for a,

$\begin{array}{c}P\left(Y>30\right)=P\left(Y\ge 30+0.5\right)\\ =P\left(Y\ge 30.5\right)\end{array}$

The formula to convert Y values into z score is,

$z=\frac{Y-np}{\sqrt{np(1-p)}}$

Substitute 25 for np and 4.3301 for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(Y\ge 30.5\right)=P\left(\frac{Y-np}{\sqrt{np(1-p)}}\ge \frac{30.5-np}{\sqrt{np(1-p)}}\right)\\ =P\left(z\ge \frac{30.5-25}{4.3301}\right)\\ =P\left(z\ge \frac{5.5}{4.3301}\right)\\ =P\left(z\ge 1.27\right)\end{array}$

The above probability can be obtained by finding the areas to the right of 1.27.

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at 1.27,

• Locate 1.2 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.07.

That is, $P\left(z\le 1.27\right)=0.8980$

Thus,

$\begin{array}{c}P\left(z\ge 1.27\right)=1-P\left(z<1.27\right)\\ =1-0.8980\\ =0.1020\end{array}$

Thus, probability that more than 30 have green seeds is 0.1020.

To determine

Find the probability that more than 30 have green seeds, out of 100 off spring of heterozygous plants.

Answer to Problem 3SE

The probability that more than 30 have green seeds is 0.1414.

Explanation of Solution

Calculation:

The general formula for the normal approximation to the binomial is,

$P\left(a\le Y\le b\right)\approx P\left(a-0.5<Y<b+0.5\right)$

Substitute 30 for a and 35 for b.

$\begin{array}{c}P\left(30\le Y\le 35\right)=P\left(30-0.5<Y<35+0.5\right)\\ =P\left(29.5<Y<35.5\right)\end{array}$

The formula to convert Y values into z score is,

$z=\frac{Y-np}{\sqrt{np(1-p)}}$

Substitute 25 for np and 4.3301 for $\sqrt{np(1-p)}$

$\begin{array}{c}P\left(29.5<Y<35.5\right)=P\left(\frac{29.5-25}{4.3301}\le z\le \frac{35.5-25}{4.3301}\right)\\ =P\left(1.04\le z\le 2.42\right)\\ =P\left(z\le 2.42\right)-P\left(z\le 1.04\right)\end{array}$

The above probability can be obtained by finding the difference between the areas to the left of 2.42 and the left of 1.04.

Use Table A.2: Cumulative Normal Distribution to find the areas to the left of z score.

Procedure:

For z at 2.42,

• Locate 2.4 in the left column of the Table V.
• Obtain the value in the corresponding row below 0.02.

That is, $P\left(z\le 2.42\right)=0.9922$

For z at 1.04,

• Locate 1.0 in the left column of the Table V.
• Obtain the value in the corresponding row below 0.04.

That is, $P\left(z\le 1.04\right)=0.8508$

Hence, the difference between the areas to the left of 2.42 and the left of 1.04 is,

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

Thus, probability that more than 30 have green seeds is 0.1414.

To determine

Find the probability that fewer than 80 have yellow seeds, out of 100 off spring of heterozygous plants.

Answer to Problem 3SE

The probability that fewer than 80 have yellow seeds is 0.8508.

Explanation of Solution

Calculation:

The random variable Y is defined as the number of plants out of 100 that have green seeds. The number of plants that have fewer than 80 yellow seeds is same as the number of plants that have more than 20 green seeds.

The approximate binomial probability using the normal distribution is,

$P\left(Y>a\right)\approx P\left(Y\ge a+0.5\right)$

By using continuity correction, the value 0.5 is added from the value of 20.

Substitute 20 for a,

$\begin{array}{c}P\left(Y>20\right)=P\left(Y\ge 20+0.5\right)\\ =P\left(Y\ge 20.5\right)\end{array}$

The formula to convert Y values into z score is,

$z=\frac{Y-np}{\sqrt{np(1-p)}}$

Substitute 25 for np and 4.3301 for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(Y\ge 20.5\right)=P\left(z\ge \frac{20.5-25}{4.3301}\right)\\ =P\left(z\ge \frac{-4.5}{4.3301}\right)\\ =P\left(z\ge -1.04\right)\end{array}$

The above probability can be obtained by finding the areas to the right of –1.04.

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at –1.04,

• Locate –1.0 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.04.

That is, $P\left(z\le -1.04\right)=0.1492$

Thus,

$\begin{array}{c}P\left(z\ge -1.04\right)=1-P\left(z<-1.04\right)\\ =1-0.1492\\ =0.8508\end{array}$

Thus, the probability that fewer than 80 have yellow seeds is 0.8508.