According to genetic theory, the blossom color in the second generation of a certain cross of sweet peas should be red or white in a 3:1 ratio. That is, each plant has probability 3/4 of having red blossoms, and the blossom colors of separate plants are independent.

a. What is the probability that exactly 6 out of 9 of these plants have red blossoms? (Use the binomial distribution. Round your answer to four decimal places.)
b. What is the mean number of red-blossomed plants when 160 plants of this type are grown from seeds? red-blossomed plants.
c. What is the probability of obtaining at least 110 red-blossomed plants when 160 plants are grown from seeds? (Use the normal approximation.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=9, p=0.75)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

And we want this probability:

[tex] P(X=6)[/tex]

And using the pmf we got:

[tex] P(X=6) = (9C6) (0.75)^6 (1-0.75)^{9-6}= 0.2336[/tex]

Part b

For this case if n = 160 the expected value would be:

[tex] E(X) = np = 160*0.75 = 120[/tex]

Part c

We need to check the conditions in order to use the normal approximation.

[tex]np=160*0.75=120 \geq 10[/tex]

[tex]n(1-p)=160*(1-0.75)=40 \geq 10[/tex]

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np=160*0.75=120[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{160*0.75(1-0.75)}=5.477[/tex]

And we want this probability:

[tex]P(X >110)[/tex]

And we can use the z score formula given by:

[tex] z =\frac{x -\mu}{\sigma}[/tex]

And replacing we got:

[tex]P(X >110)= P(Z> \frac{110-120}{5.477}) = P(Z>-1.826)= 1-P(Z<-1.826) =0.9661 [/tex]