Question

t is known that 50% of adult workers have a high school diploma. If a random sample of 5 adult workers is selected, what is the probability that more than 3 of them have a high school diploma?

Answer 1

Answer:

The probability that high school diploma is possessed by more than 3 is 0.1875.

Step-by-step explanation:

It is given that 50% of adult workers have a high school diploma.

Therefore we can write p = 0.5.

We are given to find that if a random sample of 5 adult workers is selected what is the probability that more than 3 of them have a high school diploma.

Therefore we can say that the sample size, n = 8.

This is a problem that will be solved using Binomial Theorem.

We have to find p(X > 3) which can be written as

  p(X > 3) = p(X = 4) + P(X = 5)

=  $\binom{8}{4}(0.5)^4(0.5)^1 + \binom{8}{5}(0.5)^5(0.5)^0$

= (6)($0.5^5$)

= 0.1875

The probability that high school diploma is possessed by more than 3 is 0.1875.

Answer 2

Answer:

18.75% probability that more than 3 of them have a high school diploma

Step-by-step explanation:

For each adult worker, there are only two possible outcomes. Either they have a high school diploma, or they do not. The adults are chosen at random, which means that the probability of an adult having a high school diploma is independent from 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.

It is known that 50% of adult workers have a high school diploma.

This means that $p = 0.5$

If a random sample of 5 adult workers is selected, what is the probability that more than 3 of them have a high school diploma?

This is P(X > 3) when $n = 5$

So

$P(X > 3) = P(X = 4) + P(X = 5)$

In which

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

$P(X = 4) = C_{5.4}.(0.5)^{4}.(0.5)^{1} = 0.15625$

$P(X = 5) = C_{5.5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X > 3) = P(X = 4) + P(X = 5) = 0.15625 + 0.03125 = 0.1875$

18.75% probability that more than 3 of them have a high school diploma