Question

Paul plays a video game that is scored by round . In about 500 rounds his career average is 20 points per round with a standard deviation of 5 points per round . Suppose we take random samples of past 3 rounds and calculate the mean number of points scored in each sample . What is the mean and sd of sampling distribibution

Answer 1

The mean of Paul's sampling distribution is 20 points per round, and the standard deviation is approximately 2.89 points per round, calculated using the formula for the standard error of the mean.

Step-by-step explanation:

The scenario involves a sampling distribution of the mean scores over random samples of 3 rounds from Paul's video game scores. To calculate the mean and standard deviation (sd) of the sampling distribution, we use the following formulas:

• The mean of the sampling distribution (μX) is equal to the population mean (μ), which is 20 points per round.
• The standard deviation of the sampling distribution (σX), also known as the standard error, is equal to the population standard deviation (σ) divided by the square root of the sample size (n), so σX = σ/√n. In this case, it would be 5/√3, approximately 2.89 points per round.

Therefore, the mean of the sampling distribution is 20, and the standard deviation is approximately 2.89.

Answer 2

The mean of the sampling distribution is 20 and the standard deviation is 2.89.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Population:

Mean 20, standard deviation of 5.

Sampling distribution:

3 rounds

Mean = 20

`s = (5)/(√(3)) = 2.89`

The mean of the sampling distribution is 20 and the standard deviation is 2.89.