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Fofino [41]
2 years ago
10

A chain store plans to reduce its workforce by 4%. If there are currently 18,600 employees and 126 of them retire voluntarily, h

ow many additional employees must be laid off?
PLEASE SHOW THE WORK !!!!
Mathematics
1 answer:
Alecsey [184]2 years ago
7 0
The number of additional employees who should be laid off to meet target of reducing the workforce by 4% will be as follows:
Total number of workers to be laid off will be given by:
(%ge of employees to be laid off)/100×(Total number of employees)
=4/100×18,600
=744 employees
Given that 126 of the quit voluntarily the remaining number that shpuld be laid off will be:
(total number of employees to be laid off)-(number of employees who quit)
=744-126
=618
Therefore we conclude the 618 more employees should be laid off.




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Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks
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Answer:

a) 3.75

b) 23.61% probability that fewer than 3 tanks will be found to be leaking

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Step-by-step explanation:

For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. 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.

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)}

To solve question c), i am going to approximate the binomial distribution to the normal.

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)}.

It is estimated that 25% of these tanks leak.

This means that p = 0.25

15 tanks chosen at random

This means that n = 15

a.What is the expected number of leaking tanks in such samples of 15?

E(X) = np = 15*0.25 = 3.75

b.What is the probability that fewer than 3 tanks will be found to be leaking?

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361

23.61% probability that fewer than 3 tanks will be found to be leaking

c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?

Now we have n = 2000. So

\mu = E(X) = np = 2000*0.25 = 500

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36

This probability is 1 subtracted by the pvalue of Z when X = 600. So

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

Z = \frac{600 - 500}{19.36}

Z = 5.16

Z = 5.16 has a pvalue of 0.

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