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 |
|---|---|---|---|---|---|---|
| P(x) | 0.1672 | 0.3329 | 0.2881 | 0.1488 | 0.0376 | 0.0254 |
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?
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Compute the theoretical mean of the random variable X for the given probability distribution.
[tex]\mu_x=1.633 \text { hits }[/tex]
(Round to three decimal places as needed.)
Compute the theoretical standard deviation of the random variable X for the given probability distribution.
[tex]\sigma_x=1.181 \text { hits }[/tex]
(Round to three decimal places as needed.)
Approximate the mean of the random variable X based on the simulation for 25 games.
[tex]\bar{x} \approx 1.320^7 \text { hits }[/tex]
(Round to three decimal places as needed.)
Approximate the standard deviation of the random variable X based on the simulation for 25 games.
58 [$\square$] hits
(Round to three decimal places as needed.)
Data Tables
Table of the numbers of hits for 25 games
| 2 | 4 | 1 | 0 | 1 |
|---|---|---|---|---|
| 1 | 2 | 0 | 2 | 2 |
| 1 | 1 | 0 | 1 | 1 |
| 0 | 2 | 1 | 3 | 0 |
| 0 | 4 | 2 | 0 | 2 |
Table of the numbers of hits for 50 games
| 2 | 4 | 1 | 0 | 1 | 1 | 2 | 0 | 2 | 2 |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 1 | 0 | 2 | 1 | 3 | 0 |
| 0 | 4 | 2 | 0 | 2 | 0 | 0 | 1 | 0 | 2 |
| 3 | 2 | 3 | 4 | 1 | 3 | 2 | 3 | 1 | 2 |
| 0 | 3 | 3 | 3 | 4 | 1 | 1 | 3 | 1 | 2 |
Answer (2)
Calculate the mean for 50 games: x ˉ 50 = 50 81 = 1.62 .
Calculate the standard deviation for 50 games: s 50 = 49 75.72 ≈ 1.243 .
Compare the sample mean and standard deviation to the theoretical values.
The Law of Large Numbers is illustrated, where larger sample sizes lead to values closer to the theoretical values. The approximate standard deviation of the random variable X based on the simulation for 25 games is 1.180 hits.
Explanation
Understand the problem and provided data We are given the probability distribution of the number of hits a baseball player obtained in a game. We are also given the results of simulations of 25 and 50 games. We need to approximate the mean and standard deviation of the random variable X based on the simulation for 50 games, compare the results to the theoretical mean and standard deviation, and determine what property is being illustrated.
Calculate the mean for 50 games First, we calculate the mean of the random variable X based on the simulation for 50 games. The data for 50 games is:
2, 4, 1, 0, 1, 1, 2, 0, 2, 2, 1, 1, 0, 1, 1, 0, 2, 1, 3, 0, 0, 4, 2, 0, 2, 0, 0, 1, 0, 2, 3, 2, 3, 4, 1, 3, 2, 3, 1, 2, 0, 3, 3, 3, 4, 1, 1, 3, 1, 2
The sum of these values is 81. The number of games is 50. Therefore, the sample mean is: x ˉ 50 = 50 81 = 1.62
Calculate the standard deviation for 50 games Next, we calculate the standard deviation of the random variable X based on the simulation for 50 games. The formula for the sample standard deviation is:
s = n − 1 ∑ i = 1 n ( x i − x ˉ ) 2
Where x i is the number of hits in the i -th game, x ˉ is the sample mean, and n is the number of games. We already calculated that x ˉ 50 = 1.62 . We need to calculate the sum of the squared differences from the mean:
∑ i = 1 50 ( x i − 1.62 ) 2
Using a calculator, we find that this sum is approximately 75.72. Therefore, the sample standard deviation is:
s 50 = 50 − 1 75.72 = 49 75.72 ≈ 1.5453 ≈ 1.243
Compare the sample and theoretical values We are given the theoretical mean μ x = 1.633 and the theoretical standard deviation σ x = 1.181 . We calculated the sample mean for 50 games as x ˉ 50 = 1.62 and the sample standard deviation as s 50 = 1.243 .
Comparing the sample mean and standard deviation to the theoretical values, we see that the sample mean (1.62) is very close to the theoretical mean (1.633), and the sample standard deviation (1.243) is reasonably close to the theoretical standard deviation (1.181).
Determine the property being illustrated The property being illustrated is the Law of Large Numbers. This law states that as the number of trials in a simulation increases, the sample mean and standard deviation will converge to the theoretical mean and standard deviation. In this case, we see that the sample mean and standard deviation for 50 games are closer to the theoretical values than the sample mean and standard deviation for 25 games (which was given as 1.3207 and 1.180 respectively).
State the final answer The approximate mean of the random variable X based on the simulation for 50 games is 1.62 hits, and the approximate standard deviation is 1.243 hits.
Examples
The Law of Large Numbers is a fundamental concept in statistics that has many real-world applications. For example, insurance companies use this law to predict the probability of events occurring, such as car accidents or natural disasters. By collecting data on a large number of events, insurance companies can estimate the expected number of claims they will receive and set their premiums accordingly. This principle is also used in polling and market research, where larger sample sizes lead to more accurate representations of the overall population's opinions or behaviors. In essence, the Law of Large Numbers allows us to make more reliable predictions and decisions based on data as the amount of data increases.
For the simulations of 25 games, the approximate mean was 1.440 hits and standard deviation was about 1.180 hits. For 50 games, the approximate mean was 1.620 hits and standard deviation was about 1.243 hits. These results illustrate the Law of Large Numbers, showing that larger sample sizes yield statistics closer to the theoretical values of 1.633 (mean) and 1.181 (standard deviation).
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