Academic advising: in 2014, the community college survey of student engagement reported that 32% of the students surveyed rarely or never use academic advising services. suppose that in reality, 42% of community college students rarely or never use academic advising services at their college. in a simulation we select random samples from this population. for each sample we calculate the proportion who rarely or never use academic advising services. if we repeatedly obtain random samples of 200 students, what will be the mean and standard deviation of the sampling distribution of sample proportions?

= 0.42

= 0.0349

Step-by-step explanation:

Given:

Probability that students surveyed rarely or never use academic advising services = 32% = 0.32

In reality, students rarely or never use academic advising services at their college = 42% = 0.42

Sample size, n = 200

The sampling distribution of sample proportion will be approximately normal with mean

therefore,

Mean, = p

or

= 0.42

now,

the standard deviation is given using the formula

on substituting the respective values, we get

or

= √0.001218

or

= 0.0349

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%