 # A Discrete Probability Distribution.

Essay by   •  March 20, 2019  •  Essay  •  344 Words (2 Pages)  •  186 Views

## Essay Preview: A Discrete Probability Distribution.

Report this essay
Page 1 of 2
1. A discrete probability distribution

In order to construct a discrete probability distribution, we have to choose some random variables. Let’s suppose that we have 8 fighters in the ring, we want to choose top 3 fights out of 5 at random. X will be the number of top three fights selected. X=0,1,2,3.

P(X=0)=10/56

P(X=1)=30/56

P(X=2)=15/56

P(X=3)=1/56

 x P(x) 0 10/56=0.199 1 30/56=0.536 2 15/56= 0.268 3 1/56=0.0199

Figure1 shows that the probability of no fighter selected, one, two and three fighters selected.

therefore, we have all of P(X) value above in the chart. They are adding up equal to 1. The probability distribution may look like figure 1.

The E(X)= the sum of x*P(x)=1.1317

The Variance=0.5357

Standard deviation=0.7319

One practical use for probability distributions in business is to predict future and influence decision making. We can see that predicted value is 1.1317, which is the weighted average of all possible values. Choosing 1 top 3 fight out of 8 fighters will have the greatest chance. The variance= 0.5357 means how far 0-3 are spread out from their E(x). In this case, I will have great chance to randomly select 2 top fighters out of 8 fighters to win the game for me.

1. Continuous probability distribution

Now given that salaries each job position have a bell shaped distribution.

 YEAR 1 YEAR 2 YEAR 3 Sales Manager \$23,400 \$24,570 \$25,740 Sales Staff Salaries \$35,640 \$43,290 \$66,690 Management Salary \$31,200 \$33,800 \$36,400 Financial Controller Salary \$20,000 \$21,000 \$22,000 Employer Costs (11%) \$12,126 \$13,493 \$16,591 Total Employee Benefits \$122,366 \$136,153 \$167,421

To calculate the mean and standard deviation, I have mean of column 1 is \$27560 and standard deviation is 7141.3164. I have mean of column 2 is \$ 30665 and standard deviation is 9996.334328. I have mean of column 3 is \$37707.5 and standard deviation is 20261.9633. Those information gives me an insight that there are about 68% employees in the company making salaries within the range of (27560-7141.3164, 27560+7141.3164) in year 1. there are about 95.45% employees in the company making salaries within the range of (30665 -9996.334328, 30665+9996.334328) in year 2.There are about 99.73% employees in the company making salaries within the range of (37707.5-20261.9633, 37707.5+20261.9633) in year 3.

...

...