An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 244 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.)

The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

356 dies were examined by an inspection probe and 244 of these passed the probe.

This means that

95% confidence level

So , z is the value of Z that has a p-value of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

0.6

39mph

step-by-step explanation: