Question

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. And offspring peas are randomly selected in groups of 22. Find the mean and standard deviation for the numbers of peas with green pods in groups of 22.

Answer 1

Step 1:

In this question, we are given the following:

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. And offspring peas are randomly selected in groups of 22.

Find the mean and standard deviation for the numbers of peas with green pods in groups of 22.

Step 2:

Recall that the expected value, or mean, of a binomial distribution, is calculated by multiplying the number of trials (n) by the probability of successes (p), or

`\begin{gathered} \text{n x p = np} \\ \text{where n = 22} \\ p\text{ = 0.75} \\ \end{gathered}`

Then, the mean =

`n\text{ x p = 22 x 0. 75 = 16.5}`

Therefore,

Mean = 16. 5

Step 3:

Recall that the Standard Deviation of a binomial distribution is defined as:

Hence, we have that:

`\begin{gathered} \text{Standard Deviation = }\sqrt[]{np(1-p)} \\ where\text{ n = }22, \\ p\text{ = 0. 75} \\ (\text{ 1-p ) = 1-0.75 = 0. 25} \\ \text{Hence,} \\ \text{Standard Deviation=}\sqrt[]{22\text{ x 0. 75 x ( 1-0.75) }} \end{gathered}`
`\begin{gathered} \text{Standard Deviation=}\sqrt[]{22\text{ x 0.75 x 0. 25 }} \\ =\text{ }\sqrt[]{4.125} \\ =\text{ 2.03 ( 2 decimal places)} \end{gathered}`