Question

Assume that hybridization experiments are conducted with peas having the property that for​ offspring, there is a 0.75probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 10.Complete parts​ (a) through​ (c) below.a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 10.The value of the mean is muequalsnothingpeas.

Answer 1

Answer: The mean is $\mu=7.5$ and standard deviation is $\sigma=1.37$ .

Step-by-step explanation:

The mean and standard deviation for the binomial distribution is given by :-

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

,where n is the number of trials and p is the probability of getting success in each trial.

Given : The probability that a pea has green pods : $p=0.75$

Then , the standard deviation for the numbers of peas with green pods in the groups of 10  :-

$\mu=10\times0.75=7.5\\\\\sigma=\sqrt{0.75(1-0.75)10}=1.36930639376\approx1.37$

Hence, the mean is $\mu=7.5$ and standard deviation is $\sigma=1.37$  .

Answer 2

The value of the mean and standard deviation values of the hybridization experiment is :

• Mean, μ = 7.5

• Standard deviation, σ = 1.37

Given the Parameters :

• Probability, p = 0.75

• Sample size, n = 10

To obtain the mean , we use the relation :

Mean, μ = np

Mean, μ = (10 × 0.75) = 7.5

To obtain the Standard deviation, we use the relation :

$standard \: deviation \: = \sqrt{np(1 - p)}$

$standard \: deviation \: = \sqrt{10 \times 0.75(0.25)} = 1.37$

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