Lucille has a collection of more than 500 songs on her phone that have a mean duration of 215 seconds and a standard deviation of 35 seconds. Suppose that every week she makes a playlist by taking an SRS of 49 of these songs, and we calculate the sample mean duration ë of the songs in each sample. Calculate the mean and standard deviation of the sampling distribution of . seconds L = seconds

The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.

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.

All songs

Mean 215 seconds, standard deviation 35 seconds

Sample

49

Mean 215, standard deviation

The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

since we know that the two functions are equal, f (x) = g

(x), therefore when we equate the two equations the result should be similar to

3x = x – 4:

a. f(x) = 3x, g(x) = x – 4

equating the two functions:

3x = x – 4             (correct)

so we now got our answer (a)