How can the answer to the last part of the question possibly be 27?! If it was 27, the probability used would be 0.675. With this probability, is would have been impossible to answer the binomial distribution question which went before it, as 0.675 is not on the binomial tables, whereas 0.35 is (0.35 x 40= 14). Sorry to break it to you guys, but 14 is clearly the correct answer to the last part of question 3.
The question before it had a different probability... 3 a (iv): "at least their reserve prices but not their lower price estimates" p=0.35 3 b: "at least their reserve prices but not their upper price estimates" p=0.675
How can the answer to the last part of the question possibly be 27?! If it was 27, the probability used would be 0.675. With this probability, is would have been impossible to answer the binomial distribution question which went before it, as 0.675 is not on the binomial tables, whereas 0.35 is (0.35 x 40= 14). Sorry to break it to you guys, but 14 is clearly the correct answer to the last part of question 3.
My answer was 14 -_-
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How can the answer to the last part of the question possibly be 27?! If it was 27, the probability used would be 0.675. With this probability, is would have been impossible to answer the binomial distribution question which went before it, as 0.675 is not on the binomial tables, whereas 0.35 is (0.35 x 40= 14). Sorry to break it to you guys, but 14 is clearly the correct answer to the last part of question 3.
It was 27. Your claim relies on the fact that the above events were mutually exclusive, but they weren't. An antique could fall into more than one category (e.g. Sold above reserve, sold above lower estimate). So yes, the probabilities did add up to more than one, but that's indifferent.
It was 27. Your claim relies on the fact that the above events were mutually exclusive, but they weren't. An antique could fall into more than one category (e.g. Sold above reserve, sold above lower estimate). So yes, the probabilities did add up to more than one, but that's indifferent.
I don't know about you guys, but I'm trusting the wise words of Samuel Jackson
I think for the CLT, all we had to say was that it was a normal distribution.
I put that the sample size was small i.e. n<30 So the central limit theorem didn't apply. But there was one past paper I did that didn't accept sample size as a reason
find the corresponding z value for 0.98, then use the standardising equation. And for part c, you find the z value for 0.99 (1-0.01) and put a minus sign in front of it, and then use the standardising equation again.
I think they are all right but correct me if I'm wrong
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For part c, I used the negative z value. Because I thought it didn't make sense for the probability of X<410 to be only 0.01 if the mean is 403. So the value I got was around 417.
For part c, I used the negative z value. Because I thought it didn't make sense for the probability of X<410 to be only 0.01 if the mean is 403. So the value I got was around 417.
I used the negative z value as well. Can't remember exactly the answer though...lemme quickly check.
"each bag"... I went with that meaning "all"... but I didn't know?
I get what people mean by the last question haivng to be to the power of 10 for 4 marks, however the questions states 'weight of the cement in EACH of the 10 bags@ surely the word each would mean that you would just use normal distribution?