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twenty-two percent of work injuries as back injuries are back injuries. it 400 work-injured people are selected at random, find the probability that 92 or fewer have back injuries

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If you were to divide the two you get 4.3 I believe that’s the answer

Using the normal approximation to the binomial, it is found that there is a 0.7054 = 70.54% probability that 92 or fewer have back injuries.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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, which is the percentile of X.  
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].

In this problem:

  • 22% have back injuries, hence [tex]p = 0.22[/tex].
  • 400 people are selected, hence [tex]n = 400[/tex].

For the approximation, the mean and the standard deviation are:

[tex]\mu = np = 400(0.22) = 88[/tex]

[tex]\sigma = \sqrt{np(1-p)} = \sqrt{400(0.22)(0.78)} = 8.2849[/tex]

Using continuity correction, the probability that 92 or fewer have back injuries is [tex]P(X \leq 92 + 0.5) = P(X \leq 92.5)[/tex], which is the p-value of Z when X = 92.5.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{92.5 - 88}{8.2849}[/tex]

[tex]Z = 0.54[/tex]

[tex]Z = 0.54[/tex] has a p-value of 0.7054.

0.7054 = 70.54% probability that 92 or fewer have back injuries.

To learn more about the normal approximation to the binomial, you can take a look at https://brainly.com/question/15727396

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