In a given one of Mendel’s hybridization experiments, the probability of offspri

snowlovelydayM

snowlovelydayM

Answered question

2021-09-15

In a given one of Mendel’s hybridization experiments, the probability of offspring peas having green pods was 3/4 or .75. Consider breeding a group of twelve such plants.
a.) What would be the mean and standard deviation of that binomial probability distribution?
b.) What would be the probability of seeing exactly six plants being green? Would you consider that an unusual outcome? Why or why not?
c.) What would be the probability of seeing at least nine plants being green?

Answer & Explanation

lamusesamuset

lamusesamuset

Skilled2021-09-16Added 93 answers

Step 1
Solution:
Let X be the number of plants being green and n be the number of plants.
From the given information, probability of offspring peas having green rods is 0.75 and n=12.
Here, plants are independent and probability of success is constant. Hence, X follows binomial distribution with parameters n=12 and p=0.75.
The probability mass function of binomial random variable X is
P(X=x)=(beg{array}{c}nxend{array})px(1p)nx;x=0,1,..,n
Step 2
a. Mean:
Mean=np
=12(0.75)
=9
Thus, the mean of the binomial probability distribution is 9.
Standard deviation:
Standard deviation: =np(1p)
=12(0.75)(10.75)
=1.5
Thus, the standard deviation of the binomial probability distribution is 1.5.
Step 3
b. The probability of seeing exactly six plants being green is
P(X=6)=(beg{array}{c}126end{array})0.756(10.75)126
=924×0.756×0.256
=0.0401
Thus, the probability of seeing exactly six plants being green is 0.0401.
This probability is less than 0.05. Thus, this is an unusual outcome.
Step 4
c. The probability of seeing at least nine plants being green is
P(X9)=P(X=9)+P(X=10)+P(X=11)+P(X=12)
=(beg{array}{c}129end{array})0.759(10.75)129+(beg{array}{c}1210end{array})0.7510(10.75)1210+(beg{array}{c}1211end{array})0.7511(10.75)1211+(beg{array}{c}1212end{array})0.7512(10.75)1212

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