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

0.7054

Intro Stats / AP Statistics

Chapter 6

The Normal Distribution

Section 4

The Normal Approximation to the Binomial Distribution

Probability Topics

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Lectures

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02:45

Twenty-two percent of work…

02:32

'Back Injuries Twenty…

00:44

In 2004 , the death rate p…

01:28

Twenty-five percent of all…

00:56

Suppose that 100 people en…

01:56

A local bank reports that …

Okay, problem. 15. We have an equals 400. He is 22% of it. The main is going to equal end times p and you should get 88. The standard deviation is the square root of end times P, which we said was 88 times Q, which is just one minus p. And our standard deviation is 8.28 All right, now we want to find the probability that X is less than or equal to 92. Okay, for our continuity conversion, when X is less center equal to we add 0.5 and remove the equal side. So 92 plus 0.5 gives us 92.5. That's going to be our new X value. So Z equals X minus the mean divided by the standard deviation, and the Z score is 0.54 If you look that up on the Z score table, you should get 0.7054 Okay, that's the probability that X is less than 92.5. And that's exactly what we want. Converted to percentage 70.54 Just move the decimal over to places and that's it.

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