Question 1063270: The mean salary of 350 workers at company A is $1050 and the standard deviation is $150. If we assume that the salaries are normally distributed, compute a probability,
1). a worker has salary less than $900
2). a worker has salary between $800 and $1300
3). that exist exactly 102 workers with salary more than $900
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
less than 900 is z, (900-1050)/150=-1
probability z<-1 is 0.1587
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between 800 and 1300 is z=(800-1050)/150=-250/150=-1.667 to z=(1300-1050)/150=250/150=1.667
probability -1.667 < z < 1.667=0.9044
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For salary >900, the probability is 0.8413, the complement of the first part.
We can use a binomial distribution where n=350 and p=0.8413 and (1-p)=0.1587
The calculator can do 350C102 (.8413)^102(.1587)^248=and the answer is 0.
This is not surprising since the expected value is 175 people, and if we use a binomial distribution, the variance is np(1-p) or 87.5 with a sd 9.35.
102 people is 73/9.35 = 7.80 sd s below the mean and has probability 0.
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