Mathematics, 04.04.2020 13:09 robert7248

Suppose that 60% of employees at a particular corporation participate in the optional retirement plan. If a random sample of 50 employees is selected, what is the probability that at least 25 employees in the sample will participate in the optional retirement plan?

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Answer from: cannaincorporated

94.41% probability that at least 25 employees in the sample will participate in the optional retirement plan

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

n = 50, p = 0.6

$\mu = E(X) = np = 50*0.6 = 30$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.6*0.4} = 3.46$

What is the probability that at least 25 employees in the sample will participate in the optional retirement plan?

Using continuity correction, this is $P(X \geq 25 - 0.5) = P(X \geq 24.5)$ , which is 1 subtracted by the pvalue of Z when X = 24.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24.5 - 30}{3.46}$

Z = -1.59

Z = -1.59 has a pvalue of 0.0559

1 - 0.0559 = 0.9441

94.41% probability that at least 25 employees in the sample will participate in the optional retirement plan

Answer from: Quest

not sure but it may be b, as i said i'm not sure.

Answer from: Quest

hey there! : )

4y < -4

we need to isolate y. we can do this by simply dividing both sides by 4.

4y ÷ 4 > -4 ÷ 4

simplify.

y > -1

5 - y > 0

once again, we need to isolate y, so we can start off by subtracting 5 from both sides.

5 - y - 5 > 0 - 5

simplify.

-y > -5

divide both sides by -1.

** remember that when dividing any side by a negative, you must switch the sign! **

-y ÷ -1 < -5 ÷ -1

simplify.

y < 5

~hope i ! ~

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