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).

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this