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

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

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

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

5

step-by-step explanation: