Mathematics, 26.06.2020 16:01, rrusso4

It is known that 50% of adult workers have a high school diploma. If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma

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answered: babyskitt

85.56% probability that less than 6 of them have a high school diploma

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have a high school diploma, or they do not. The probability of an adult having a high school diploma is independent of other adults. 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.

50% of adult workers have a high school diploma.

This means that p = 0.5

If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma

This is P(X < 6) when n = 8.

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

In which

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

$P(X = 0) = C_{8,0}.(0.5)^{0}.(0.5)^{8} = 0.0039$

$P(X = 1) = C_{8,1}.(0.5)^{1}.(0.5)^{7} = 0.0313$

$P(X = 2) = C_{8,2}.(0.5)^{2}.(0.5)^{6} = 0.1094$

$P(X = 3) = C_{8,3}.(0.5)^{3}.(0.5)^{5} = 0.2188$

$P(X = 4) = C_{8,4}.(0.5)^{4}.(0.5)^{4} = 0.2734$

$P(X = 5) = C_{8,5}.(0.5)^{5}.(0.5)^{3} = 0.2188$

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0039 + 0.0313 + 0.1094 + 0.2188 + 0.2734 + 0.2188 = 0.8556

85.56% probability that less than 6 of them have a high school diploma

answered: Guest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

answered: Guest

1 and 0

step-by-step explanation:

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