Mathematics, 20.06.2020 17:57 tyliyahmiles99

A real estate agent has 17 properties that she shows. She feels that there is a 40% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling less than 5 properties in one week. Round your answer to four decimal places.

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Answer from: yeet1570

0.126 = 12.6% probability of selling less than 5 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

A real estate agent has 17 properties that she shows.

This means that n = 17

She feels that there is a 40% chance of selling any one property during a week.

This means that p = 0.4

Compute the probability of selling less than 5 properties in one week.

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

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

$P(X = 0) = C_{17,0}.(0.4)^{0}.(0.6)^{17} = 0.0002$

$P(X = 1) = C_{17,1}.(0.4)^{1}.(0.6)^{16} = 0.0019$

$P(X = 2) = C_{17,2}.(0.4)^{2}.(0.6)^{15} = 0.0102$

$P(X = 3) = C_{17,3}.(0.4)^{3}.(0.6)^{14} = 0.0341$

$P(X = 4) = C_{17,4}.(0.4)^{4}.(0.6)^{13} = 0.0796$

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0019 + 0.0102 + 0.0341 + 0.0796 = 0.126

0.126 = 12.6% probability of selling less than 5 properties in one week.

Answer from: Quest

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.