Respuesta :
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]p = 0.08, n = 1000[/tex]
So
[tex]\mu = E(X) = np = 1000*0.08 = 80[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.08*0.92} = 8.58[/tex]
What is the probability that more than 70 workers will meet with an accident during the 1-yr period
Using continuity correction, this is [tex]P(X \geq 70 + 0.5) = P(X \geq 70.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 70.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70.5 - 80}{8.58}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] 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
The probability that more than 70 workers will be involved in an accident is 0.8665
The given parameters are:
[tex]\mathbf{n = 1000}[/tex] --- population
[tex]\mathbf{p = 0.08}[/tex] --- the probability that a worker meets an accident
[tex]\mathbf{x = 70}[/tex] -- the number of workers
Start by calculating the mean and the standard deviation
[tex]\mathbf{\mu = np}[/tex] --- mean
So, we have:
[tex]\mathbf{\mu = 1000 \times 0.08}[/tex]
[tex]\mathbf{\mu = 80}[/tex]
[tex]\mathbf{\sigma = \sqrt{\mu(1 - p)}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{80 \times (1 - 0.08)}}[/tex]
[tex]\mathbf{\sigma = \sqrt{73.6}}[/tex]
[tex]\mathbf{\sigma = 8.58}[/tex]
The probability is then represented as
[tex]\mathbf{P(x > 70) = P(x > 70.5)}[/tex] ---- By continuity correction
Calculate the z-score for x = 70.5
[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]
So, we have:
[tex]\mathbf{z = \frac{70.5 - 80}{8.58}}[/tex]
[tex]\mathbf{z = -1.11}[/tex]
So, we have:
[tex]\mathbf{P(x > 70) = P(z > -1.11)}[/tex]
Using z-scores of probabilities, we have:
[tex]\mathbf{P(x > 70) = 0.8665}[/tex]
Hence, the probability that more than 70 workers will be involved in an accident is 0.8665
Read more about probabilities at:
https://brainly.com/question/15992180
