Question

2) Consider a random variable with the following probability distribution: P(X = 0) = 0.1, P(X=1) =0.2, P(X=2) = 0.3, P(X=3) = 0.3, and P(X=4)= 0.1.

A. Generate 400 values of this random variable with the given probability distribution using simulation.

B. Compare the distribution of simulated values to the given probability distribution. Is the simulated distribution indicative of the given probability distribution? Explain why or why not.

C. Compute the mean and standard deviation of the distribution of simulated values. How do these summary measures compare to the mean and standard deviation of the given probability distribution?

Answer 1

Given Data is :

a random variable with the following probability distribution: P(X = 0) = 0.1, P(X=1) =0.2, P(X=2) = 0.3, P(X=3) = 0.3, and P(X=4)= 0.1

1.First create the table as following:

Count Simulated Probab lity Umits Cumulative Probabiy Probability 100 107 107 GO If a number lies between 30 and G0 let be 2 0 If a number lies between 60 and 90 let be 3 00 If a mumber lies between 90 and 100 let be Total

2. Create a new column by multiplying probability into 100

3. Find the cumulative total of the multiplied probability. (Formula is shown)
4. Now we need to create limits to assign the numbers simulated to the number reported. This is shown in the Limit column.
Explanation: Let us generate a random number between 1 to 100. The probability of any number being generated is 0.01.
Total probability of a number being generated to less than or equal to 50 is 0.5(Addition of 50 independent probability). Using this logic the probability will be simulated.
5.Generate 400 random number between 1 to 100 as shown below.
6. Assign them the simulated numbers as per the limit logic we have done before.
7. Find the count of each simulated reported number as shown in the table.
8. We then find the simulated probability, count/total number.

random numberSimulated Number reported random number RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) RANDBETWEEN(1,100) -RANDBETWEEN(1,100) RANDBETIN EEN(1 100) 0 0 4 0

  We see that the distribution of simulated values is indicative of given probability distribution. There is a slight deviation from the probability due to sampling error.

The mean of the original distribution is shown below :

(1-μ)2 x P(z) (x-μ)2 4.41 1.21 0.01 0.81 .61 0.441 0.242 0.003 0.2 0.3 0.3 0.2 0.6 0.9 0.4 0.361 Mean srt(1.29) 1.135781669 Standard deviation

The mean of the simulated distribution is shown below:

PIx) rx P(X) 4.3472 1.17722 0.0072 0.837 3.65722 0.0925 0.2475 0.2675 0.2675 0.125 0,402118313 0.211363188 0.001932689 0.223957688 0.458403125 0.2475 0.535 0.8025 0.5 Σ(x-p)2 × P(z) 1.377775 Mean -SORT( 1.3끼 1.173785608 Standard deviation

and the MEAN for both are equal.

THANK YOU..