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

last one.

step-by-step explanation:

last one.

generally speaking, the others follow some sort of pattern that correlates to either linear (looks like a straight line) or sinusoidal (looks like an s).

the last one is just a random cloud of points that doesn't correlate to any particular known order.

Answer from: Quest

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

Answer the following and be the brainiest the mistake in the pattern is that : a. the decimal point

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Suppose that 60% of employees at a particular corporation participate in the optional retirement pla...