Question 5: A recent CNN News survey reported that 76% of adults think the U.S. pennies should still be made. Suppose we select a sample of 20 people.

How many of the 20 would you expect to indicate that the Treasury should continue making pennies?

What is the standard deviation?

What is the likelihood that exactly eight people would indicate the Treasury should continue making pennies?

What is the likelihood that 10 to 15 adults would indicate the Treasury should continue making pennies?

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

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

[tex]X \sim Binom(n=20, p=0.76)[/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]

Part a

For this case the expected value for the binomial distribution is given by:

[tex] E(X) =np = 20*0.76=15.2[/tex]

Part b

The variance for the binomial distribution is given by:

[tex] Var(X) = np(1-p) = 20*0.76*(1-0.76) =3.648[/tex]

And the deviation would be ust the square root of the variance and we got:

[tex] Sd(X) = \sqrt{3.648}=1.910[/tex]

Part c

For this case we want this probability:

[tex]P(X=8)=(20C8)(0.76)^8 (1-0.76)^{20-8}=0.000512[/tex]

That correspond to approximately 0.0512%, so then we can conclude that is very unlikely since is <1%

Part d

For this case we want this probability:

[tex] P(10 \leq X \leq 15)=P(X=10)+....+P(X=15)[/tex]

If we find the individual probabilities we got:

[tex]P(X=10)=(20C10)(0.76)^{10} (1-0.76)^{20-10}=0.0075[/tex]

[tex]P(X=11)=(20C11)(0.76)^{11} (1-0.76)^{20-11}=0.0217[/tex]

[tex]P(X=12)=(20C12)(0.76)^{12} (1-0.76)^{20-12}=0.0515[/tex]

[tex]P(X=13)=(20C13)(0.76)^{13} (1-0.76)^{20-13}=0.100[/tex]

[tex]P(X=14)=(20C14)(0.76)^{14} (1-0.76)^{20-14}=0.159[/tex]

[tex]P(X=15)=(20C15)(0.76)^{15} (1-0.76)^{20-15}=0.201[/tex]

And if we add the values we got:

[tex] P(10 \leq X \leq 15)=0.541[/tex]