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. Assume that the offspring peas are randomly selected in groups of 22.Find the mean and standard deviation for the numbers of peas with green pods in the groups of 22.

Answer 1

Given:

`\begin{gathered} p=0.75 \\ n=22 \\ \mu=? \\ \sigma=? \end{gathered}`

Using the formula for the mean and standard deviation of binomial distribution;

`\begin{gathered} \mu=np \\ \text{where;} \\ \mu\text{ is the mean} \\ n\text{ is the number of trials} \\ p\text{ is the probability of success} \\ \\ \\ \text{Also,} \\ \sigma=\sqrt[]{npq} \\ \text{where;} \\ \sigma\text{ is the standard deviation } \\ n\text{ is the number of trials} \\ p\text{ is the probability of success} \\ q\text{ is the probability of failure} \end{gathered}`

Hence, the mean is;

`\begin{gathered} \mu=np \\ \mu=22*0.75 \\ \mu=16.5 \end{gathered}`

Therefore, the mean is 16.5

The standard deviation is;

`\begin{gathered} \sigma=\sqrt[]{npq} \\ n=22 \\ p=0.75 \\ q=1-0.75=0.25 \\ \\ \sigma=\sqrt[]{22*0.75*0.25} \\ \sigma=\sqrt[]{4.125} \\ \sigma=2.031 \end{gathered}`

Therefore, the standard deviation is 2.031