Apopulation of cats has a population mean of μ = 12μ=12pounds and a population standard deviation of σ = 2.0σ=2.0pounds. the distribution of weights of these cats is fairly symmetrical. if you take a random sample of 100 cats from this population, what will be the resulting sampling distribution of x?

Normal (12, 0.20)

Step-by-step explanation:

Given that a population of cats has a population mean of

Also given that the distribution of weights of these cats is fairly symmetrical.

Since sample size is large and a random sample of 100 cats are drawn from this population, we find that

the mean of the sample would follow a normal distribution with mean =population mean and std deviation = population std dev/sqrt n

i.e. Sampling distribution of X would be normal with

mean = 12 pounds

and std deviation =

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."