The probability distribution of the random variable X represents the number of hits a baseball player obtained in a game for the 2012 baseball season. x 0 1 2 3 4 5 Upper P (x )0.1664 0.3302 0.2933 0.1485 0.0369 0.0247 The probability distribution was used along with statistical software to simulate 25 repetitions of the experiment​ (25 games). The number of hits was recorded. Approximate the mean and standard deviation of the random variable X based on the simulation. The simulation was repeated by performing 50 repetitions of the experiment. Approximate the mean and standard deviation of the random variable. Compare your results to the theoretical mean and standard deviation. What property is being​ illustrated?

The table below shows the number of hits for 25 and 50 repetitions.

Answer and Step-by-step explanation: Expected Value or Mean of a probability distribution is calculated as:

E(X) = μ = ∑xP(X)

In a game for the 2012 baseball season, expected value is:

[tex]\mu=0(0.1670)+1(0.3320)+2(0.2864)+3(0.1499)+4(0.0381)+5(0.0266)[/tex]

μ = 1.640

Standard Deviation (σ) of a probability distribution is:

[tex]\sigma=\sqrt{\Sigma (x-\mu)^{2}P(X)}[/tex]

[tex]\sigma = \sqrt{(0-1.640)^{2}(0.1670)+...+(5-1.640)^{2}(0.0266)}[/tex]

σ = 1.1883

The simulate experiment of 25 repetitions shows a table. With this table, it is possible to create a frequency table as follows:

x    |    f

0   |    4

1    |    7

2   |   11

3   |   2

4   |    1

With a frequency table, expected value is calculated as:

[tex]\mu=\frac{\Sigma (fx)}{\Sigma f}[/tex]

[tex]\mu=\frac{0*4+1*7+2*11+3*2+4*1}{25}[/tex]

μ = 1.56

Standard deviation, using a frequency table, is calculated as:

[tex]\sigma = \sqrt{\frac{n\Sigma (fx^{2})-(\Sigma fx)^{2}}{n(n-1)} }[/tex]

[tex]\sigma=\sqrt{\frac{25(0*4^{2}+...+4*1^{2})-(0*4+...+4*1)^{2}}{25(24)} }[/tex]

σ = 0.983

The same can be done for 50 repetitions.

The frequency table is

x      |      f

0     |      6

1      |     18

2     |     16

3     |     9

4     |     1

[tex]\mu=\frac{0*6+1*18+2*16+3*9+4*1}{50}[/tex]

μ = 1.62

[tex]\sigma=\sqrt{\frac{50(0*6^{2}+...+4*1^{2})-(0*6+...4*1)^{2}}{50(49)} }[/tex]

σ = 0.9775

Comparing expected values and standard deviations, observe that the bigger the repetition of the experiment, mean and standard deviation get closer to the theoretical statistical values. This is an example of the Law of Large Numbers, a theorem stating as the number of random, normally distributed values increase, their sample mean approaches the theoretical mean.

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