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 randomly sample 200 students from this population repeatedly, the standard error is approximately 3.5%. Is it unusual to see 32% who rarely or never use academic advising services in one of these samples

means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Z scores below -2 are considered unusually low.

Z scores above 2 are considered unusually high.

For a sample proportion p in a sample of size n, we have that

In this problem, we have that:

Is it unusual to see 32% who rarely or never use academic advising services in one of these samples

What is the z-score for X = 0.32?

means that it would be unusual to see 32% who rarely or never use academic advising services in one of these samples

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: