Asurvey of 1000 air travelers1 found that 60 % prefer a window seat. the sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is startitalic upperword s e enditalic equals 0.015.

use a normal distribution to find a 90 % confidence interval for the proportion of air travelers who prefer a window seat.

round your answers to three decimal places.

the 90% confidence interval is

The 90% confidence interval is 0.575 to 0.625.

Step-by-step explanation:

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

In which

Z is the zscore that has a pvalue of .

For this problem, we have that:

90% confidence interval

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

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval is 0.575 to 0.625.

answer:

a=9

step-by-step explanation: lmk if you need a explanation

$119.20

step-by-step explanation:

$372.50 - 60% off is $149. $149-20% off is $119.20