Question

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

Answer 1

Answer:

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:

$p = 0.08, n = 1000$

So

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

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

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