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Looking to do this with calculator do I use the normalcdf ? I came up with...

Looking to do this with calculator do I use the normalcdf ? I came up with fro the first one .1642
2.) Each American uses 650 pounds of paper a year. Suppose that the distribution is approximately normal with a population standard deviation of 153.5 pounds. Find the probability that a randomly selected American uses: a.) More than 800 pounds a year-normalcdf( b.) Less than 400 pounds a year c.) Between 500 and 700 pounds a year 3.) The average number of available jobs for registered nurses is 103,900. If we assume a normal distribution with a standard deviation of 8040, find the probability that: a.) More than 100,000 jobs are available for RNs. b.) More than 80,000 jobs but less than 95,000 jobs are available for RNs. c.) If the probability is 0.1977 that more than X amount of jobs are available, find the value of X.
2.) Each American uses 650 pounds of paper a year. Suppose that the distribution is approximately normal with a population standard deviation of 153.5 pounds. Find the probability that a randomly selected American uses: a.) More than 800 pounds a year-normalcdf( b.) Less than 400 pounds a year c.) Between 500 and 700 pounds a year 3.) The average number of available jobs for registered nurses is 103,900. If we assume a normal distribution with a standard deviation of 8040, find the probability that: a.) More than 100,000 jobs are available for RNs. b.) More than 80,000 jobs but less than 95,000 jobs are available for RNs. c.) If the probability is 0.1977 that more than X amount of jobs are available, find the value of X.
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Answer #1

2. Here distribution is normal with mean=650 and standard deviation=153.5

a. We need to find P( > 800)

As distribution is normal we can convert x to z

P(x>800)=P(z>\frac{800-650}{153.5})=P(z>0.98)=0.1635

b. P(x<400)=P(z<\frac{400-650}{153.5})=P(z<-1.63)=0.0516

c. P(500<x<700)=P(\frac{500-650}{153.5}<z<\frac{700-650}{153.5})=P(-0.98<z<0.33)=0.4658

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