Tutoring Services: The Community College Survey of Student Engagement reports that 46% of the students surveyed rarely or never use peer or other tutoring resources. Suppose that in reality 40% of community college students never use tutoring services available at their college. In a simulation we select random samples from a population in which 40% do not use tutoring. For each sample we calculate the proportion who do not use tutoring. If we randomly sample 500 students at a time, what will be the mean and standard error of the sampling distribution of sample proportions? Mean: (Round to two decimal places.) Standard error: (Round to exactly three decimal places.)

Mean: 0.400

Standard error: 0.022

Step-by-step explanation:

We are taking samples of size n=500 out of a population with parameter p=0.40.

The expected distribution is the sampling distribution of sampled proportions. This distribution has parameters that are calculated as:

Mean: the mean of the sampling distribution is equal to the population proportion, as it is not biased.

In this case, the mean of this sampling distribution is p=0.40.

Standard error: the standard error depends on the population proportion and the sample size. It is calculated as:

being p: population proportion, N: sample size.

if farmer matt works for 8 hours a day, how many days will it take him to completely plant 12 rows of crops? each batch of cookies needs 3/4 cups of sugar.

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.