Question

see attachment

Answer 1

hey but dont u have to do this for the standard deviation: \[\sigma=\frac{ 3.29 }{ \sqrt{3} } since they took random sample of three waiters\]

Answer 2

@bambimonster You're correct =] I forgot about that. Correcting myself, the new z-score would be:\[\bf z=\frac{ x-\mu }{ \sigma }=\frac{ 10.23-6.66 }{ 1.8995 } \approx 1.88\]Now we look at the z-score table to calculate the probability and we get:\[\bf P(Z<1.88)=0.9699\]

Thank you for correcting me and the answer would be a.) =]

@bambimonster

Answer 3

Once again, this is only a normal approximation since the distributed is skewed to the right.

Answer 4

hey i submitted a before, and they graded me wrong, thats why i posted the question here. can it be no then?

Answer 5

Well it should be a.) but if it's not then either the question is wrong or the answer is d.).

Answer 6

okay thanks alot! (b) If we take a random sample of 35 of the waiter's customers, what is the probability that the mean tip for these customers is greater than $7.62?
i got 0.0418

Answer 7

\[\frac{ 7.62-6.66 }{ \frac{ 3.29 }{ \sqrt{35} } }=1.726 so its 1-0.9582=0.0418\]

Answer 8

Using my standard normal distribution table I get: P = 0.0422

Answer 9

did u use 1.72? or 1.73.. i used 1.73 since its 1.726

Answer 10

I used 1.726. My table gives a result for that, being 0.9573 +0.0005 = 0.9578

Answer 11

oh okay.. because im using the standard normal probabilities chart given by my teacher.. and they give only up to 2 dp.. but should i take 1.72 or 1.73..

Answer 12

If you take the average of the results for 1.72 and 1.73 you get 0.9578.

Answer 13

okay. thanks alot.. as for the first question, do u think its no? my first attempt answer was a, and i was graded wrong for it..

Answer 14

Well, I get answer a as well for the first part.

Answer 15

Presumably the sample size is too small to enable a valid calculation. So 'no' must be the answer.