Elem Statistics: discrete probability and binomial distribution

Question

52% of all Americans are home owners. If 45 Americans are randomly selected, find the probability that:

a) exactly 25 of them are home owners

b) at most 22 of them are home owners

c) at least 24 of them are home owners

d) between 22 and 29 (including 22 and 29) of them are home owners.

Answer 1

So yes, the underlying concept is binomial experiment (since it fits the four criteria, like how there are two outcomes - "success"(X) = are homeowners, "failure" = are not homeowners - and there's a fixed number of trials), and thus we'll use the following binomial probability formula:

P(X = x) = [n! / (n-X)!*X!] * p^X * (1-p)^n-X

(Note that you can simplify different parts of the formula to help with typing this formula into the calculator. Also, 1-p is also referred to as the simple variable "q" to represent the complement value to the probability of success, "p".)

Now, given the general information of p = 0.52 and n = 45 related to the variable of interest, X being "Americans who are (or are not) homeowners", let's consider each value to find their probabilities:

(a) Find the probability that exactly 25 (out of 45 Americans) are homeowners

"exactly 25" can be translated into the mathematical statement, X = 25. Thus we only need to use the binomial probability formula once for this single value:

P(X = 25) = [45! / (45-25)!*25!] * (.52)²⁵ * (1-.52)^45-25

when simplifying different parts of the formula...

P(X = 25) = [45! / 20!*25!] * (.52)²⁵ * (.48)²⁰ = .106157 (round accordingly; typically out to four decimal places)

If you're using the TI-84 calculator (and this function can be similarly found in other technologies/calculators), you can use the built-in functions binompdf (probability density function, for exact values) and binomcdf (cumulative density function, for interval/range of values "up to" a boundary value).

Like so: binompdf(n, p, X) --> P(X = 25) = binompdf(45, .52, 25) = .106157

(b) Find the probability that at most 22 are homeowners

This is where the calculator built-in function can be soo helpful because in this part, you would have to evaluate the formula twenty-two times for each value going up to 22 then add it all up, like so:

P( X </= 22) = P(0) + P(1) + P(2) + P(3)....+ P(22)

So instead, let's use the binomcdf function: P(X </= 22) = binomcdf(45, .52, 22) = .3936

(c) Find the probability that at least 24 are homeowners

To make things simpler, let's use the complementary events concept for "at least" cases, because we can find with the binomcdf function the probability up to 23 homeowners and then subtract that from 1 to get the other interval from 24 and above:

P(X >/= 24) = 1 - P(X </= 23) = 1 - binomcdf(45, .52, 23) = .4889

(d) Find the probability that between 22 and 29 are homeowners

P(22 </= X </= 29) = P(X </= 29) - P(X </= 22) = binomcdf(45, .52, 29) - binomcdf(45, .52, 22) = .9666 - .3936 = .5730

Hope you find this helpful! Peace.