A certain type of golfball is tested by a robot that hits the ball with a consistent impact force. The distances this type of ball travels in this test are normally distributed with a mean of 195 meters and a standard deviation of 5 meters. Suppose that quality control experts regularly collect random samples of 25 distance measurements and calculate the sample mean distance in each sample. Assume that the measurements in each sample are independent. What will be the shape of the sampling distribution of the sample mean distance?
Skewed to the left
(Choice A)
A
Skewed to the left

(Choice B)
B
Skewed to the right

(Choice C)
C
Approximately normal

(Choice D)
D
Unknown; we don't have enough information to determine the shape

C Approximately normal

Step-by-step explanation:

Central Limit Theorem

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.

In this question:

Sampling distribution of the sample mean is approximately normal, by the Central Limit Theorem. The correct answer is given by Choice C.

the amount you'd have after the work is $1.55

step-by-step explanation:

you subtract 95% from $31.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)