A PGA (Professional Golf Association) tournament organizer is attempting to estimate the average number of strokes for the 13th hole on a given golf course. On a particular day, 64 players completed the play on the 13th hole, with an average of 4.25 strokes and a population standard deviation of 1.6 strokes. Determine the 95 percent confidence interval for the average number of strokes.

The 95 percent confidence interval for the average number of strokes is (3.858, 3.642).

Step-by-step explanation:

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 .

So it is z with a pvalue of , so

Now, find 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 mean subtracted by M. So it is 4.25 - 0.392 = 3.858

The upper end of the interval is the mean added to M. So it is 4.25 + 0.392 = 4.642

The 95 percent confidence interval for the average number of strokes is (3.858, 3.642).

your answer is c. 97 (don't delete this again)

step-by-step explanation:

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %