Acompany makes auto batteries. they claim that 86% of their ll70 batteries are good for 70 months or longer. assume that this claim is true. let p be the proportion in a random sample of 80 such batteries for a populations that are good for 70 months or more.
what is the probability that this sample proportion is within 0.03 of the population proportion?

The probability that the sample proportion of 80 LL70 batteries is within 0.03 of the population proportion is 0.44

Step-by-step explanation:

Sample proportion being within margin, or margin of error (ME) around the mean can be found using the formula

ME= where

Then 0.03= from this we get:

z≈0.773 and the p(z)≈0.439

Therefore, the probability that the sample proportion is within 0.03 of the population proportion is 0.44

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points