Question

can someone check my answers for this problem

Answer 1

yes

Answer 2

Your mail-order company advertises that it ships 95% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that only 91 of these orders were shipped on time.

A) What is the sample proportion of orders shipped on time?

B) If the company really ships 95% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller? (Use a normal approximation, but be sure to justify the standard deviation and approximation first.)

C) A critic says, "Aha! You claim 95% but in your sample the on-time percentage is lower than that. So the 95% claim is wrong." Does your probability calculation in (b) support or refute the 95% claim? Explain.

Answer 3

A) 0.91


B) P(≤91.5)=P(z≤91.5-95/2.179)=p(z≤-1.6)=.054

C) If it was less than 5% we could not support that claim. That's not likely to happen in a population with 95 % true shipping performance. In this case 5.4% so we can probably support the claim.

Answer 4

I don't see any issues, they're all right for me (:

Answer 5

@ganeshie8 can you check my answers for this problem ?

Answer 6

did u use normal approximation to binomial theorem ?

Answer 7

\(\large mean = 100 \times 0.95 = 95\)

\(\large standard~dev = \sqrt{100 \times 0.95\times 0.05} = 2.179 \)

Answer 8

All parts are right !
And yes, since the probability of "<= 91" is greater than 5%, we cannot reject the company's claim. So we reject the critic.