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.

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15