Suppose you want to estimate the proportion of cars that are sport utility vehicles (SUVs) being driven in Kansas City, Missouri, at rush hour by standing on the corner of I-70 and I-470 and counting SUVs. You believe the figure is no higher than 0.40. If you want the error of the confidence interval to be no greater than .03, how many cars should you randomly sample? Use a 90% level of confidence.

The sample size 'n' = 721

Number of cars 'n' = 721

 Step-by-step explanation:

Explanation:-

Given  Population proportion 'p' = 0.40

Given Margin of error = 0.03

90% of level of significance

The Margin of error is determined by

on calculation , we get

0.03√n = 0.8058

squaring on both sides , we get

n = 721.45≅721

conclusion:-

The sample size 'n' = 721

Number of cars 'n' = 721

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%