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

7 0

3 years ago

6 more than 3 times what number squares is 81

Lisa [10]

When you write that equation, you get 3x^2+6=81.

Solving for x:

Subtract by 6 on both sides of the equation to get 3x^2=75.

Then, divide by 3 on both sides of the equation to get x^2=25.

Finally, do the square root on both sides of the equation to get x=5.

The number is 5.

6 0

3 years ago