Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enrolled in. The students collect a random sample of 47 students. The mean of the sample is 12.3 units. The sample has a standard deviation of 1.9 units. What is the 95% confidence interval for the average number of units that students in their college are enrolled in

The 95% confidence interval for the average number of units that students in their college are enrolled in is :

Confidence Interval ( 11.76, 12.84).

Step-by-step explanation:

The formula for a Confidence Interval is:

C. I = μ ± z × σ/√n

Where

z = z score

μ is the sample mean

σ is the sample standard deviation

n = number of samples

We were given a 95% confidence interval

The z score for a 95% confidence interval = 1.96

μ = 12.3 units

σ = 1.9

n = 47 students

C. I = μ ± z × σ/√n

C.I = 12.3 ± 1.96 × 1.9/√47

C.I = 12.3 ± 0.5432012283

Hence,

Confidence interval = 12.3 ± 0.5432012283

12.3 - 0.5432012283 = 11.756798772 Approximately ≈ 11.76

12.3 + 0.5432012283 = 12.843201228

Approximately ≈ 12.84

Therefore, the 95% confidence interval for the average number of units that students in their college are enrolled in is :

Confidence Interval ( 11.76, 12.84).

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.