A simple random sample of size 51 has sample mean of 70.72. The population distribution is approximately normal, with population standard deviation of 16.25. Determine the correct method of finding a 95% confidence interval for the unknown population mean and compute it. z-method: (66.26, 75.18) Cannot compute: the population size is too small. t-method: (66.15, 75.29) z-method: (66.15, 75.29)

z-method: (66.26, 75.18)

Step-by-step explanation:

We have the standard deviation for the population, which means that we use the z-distribution to solve this question.

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 1.96.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 70.72 - 4.46 = 66.26

The upper end of the interval is the sample mean added to M. So it is 70.72 + 4.46 = 75.18

The correct answer is z-method: (66.26, 75.18)

answer is x<

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x