Mathematics, 21.04.2020 17:56 82coogs

The Colorado Mining and Mineral Company has 1000 employees engaged in its mining operations. It has been estimated that the probability of a worker meeting with an accident during a 1-yr period is 0.08. What is the probability that more than 70 workers will meet with an accident during the 1-yr period

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Answer from: soapai

The probability that more than 70 workers will be involved in an accident is 0.8665

The given parameters are:

$\mathbf{n = 1000}$ --- population

$\mathbf{p = 0.08}$ --- the probability that a worker meets an accident

$\mathbf{x = 70}$ -- the number of workers

Start by calculating the mean and the standard deviation

$\mathbf{\mu = np}$ --- mean

So, we have:

$\mathbf{\mu = 1000 \times 0.08}$

$\mathbf{\mu = 80}$

$\mathbf{\sigma = \sqrt{\mu(1 - p)}}$

So, we have:

$\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}$

$\mathbf{\sigma = \sqrt{73.6}}$

$\mathbf{\sigma = 8.58}$

The probability is then represented as

$\mathbf{P(x 70) = P(x 70.5)}$ ---- By continuity correction

Calculate the z-score for x = 70.5

$\mathbf{z = \frac{x - \mu}{\sigma}}$

So, we have:

$\mathbf{z = \frac{70.5 - 80}{8.58}}$

$\mathbf{z = -1.11}$

So, we have:

$\mathbf{P(x 70) = P(z -1.11)}$

Using z-scores of probabilities, we have:

$\mathbf{P(x 70) = 0.8665}$

Hence, the probability that more than 70 workers will be involved in an accident is 0.8665

Read more about probabilities at:

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Answer from: 10040816

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

Step-by-step explanation:

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:

p = 0.08, n = 1000

$\mu = E(X) = np = 1000*0.08 = 80$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58$

What is the probability that more than 70 workers will meet with an accident during the 1-yr period

Using continuity correction, this is $P(X \geq 70 + 0.5) = P(X \geq 70.5)$ , which is 1 subtracted by the pvalue of Z when X = 70.5. So

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

$Z = \frac{70.5 - 80}{8.58}$

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

86.65% probability that more than 70 workers will meet with an accident during the 1-yr period

Answer from: Quest

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Answer from: Quest

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