We run a large factory with many employees. To operate at peak efficiency, we ne

Question

We run a large factory with many employees. To operate at peak efficiency, we need exactly 150 people to show up to work. Below is a table of probability of a certain number of people not showing up to work. How many absences do we expect each day? What is the standard deviation of absences each day? What is the probability of having 7 or more absent employees on a given day?

Absences Odds

0 2.41%

1 6.78%

2 12.99%

3 16.48%

4 18.21%

5 16.33%

6 12.48%

7 7.29%

8 5.47%

9 1.44%

10 0.12%

Absences Odds

0 2.41%

1 6.78%

2 12.99%

3 16.48%

4 18.21%

5 16.33%

6 12.48%

7 7.29%

8 5.47%

9 1.44%

10 0.12%

Explanation / Answer

f=   1  
fx =   4.2052  
      
Mean = fx / f =   4.2052  
Mean square = f x^2 / f =   21.9188  
      
Varriance = (Mean square) - (Mean)^2      
      
Varriance = f x^2 - Mean^2 =   4.235  
Stadard Dev= Var =   2.058  

c.

probability of having 7 or more absent employees on a given day = P(X>=7) = P(X=7) + P(X=8) + P(X=9) + P(X=10) = 0.0729 + 0.0547 + 0.0144 + 0.0012 = 0.1432

Values ( X ) Frequency(f) fx ( X^2) f x^2 0 0.0241 0 0 0 1 0.0678 0.0678 1 0.0678 2 0.1299 0.2598 4 0.5196 3 0.1648 0.4944 9 1.4832 4 0.1821 0.7284 16 2.9136 5 0.1633 0.8165 25 4.0825 6 0.1248 0.7488 36 4.4928 7 0.0729 0.5103 49 3.5721 8 0.0547 0.4376 64 3.5008 9 0.0144 0.1296 81 1.1664 10 0.0012 0.012 100 0.12

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