A hair stylist has been in business one year. Fifty percent of his customers are walk-in business. If he randomly samples nine of the people from last week’s list of customers, what is the probability that three or fewer were walk-ins? If this outcome actually occurred, what would be some of the explanations for it?

The probability that three or fewer were walk-ins is P=0.254.

If this outcome occured, it can be produced by pure chance, as it nos a very unlikely event, although the expected number of walkins is 4.5.

Step-by-step explanation:

This problem can be modeled as a binomial experiment, with p=0.5 and n=9.

p is the probability that the customers are walk-ins.

n is the sample size.

We have to calculate the probabilty that three or fewer are walk-ins.

step-by-step explanation:

from the graph attached below:

domain of graph is the value of x for which the given function profit function p(x) is defined.

(a) the domain represents the number of years for which the outlet will earn a profit.

range of profit function defines the lowest to highest value of profit earned.

(b) the range of this function is (0 , 160 ). for this function, the range represents the outlet's profit earned.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!