Question

Suppose that 37% of college students own cats. If you were to ask random college students if they own a cat what would the probability that:
a) a single student doesn’t own a cat?
b) 3 students own cats?
c) 2 students own cats while 2 students don't own cats?

Answer 1

Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37$

Item b:

The probability is P(X = 3) when n = 3, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065$

Item c:

The probability is P(X = 2) when n = 4, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260$

More can be learned about the binomial distribution at brainly.com/question/24863377

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