Given P(A)=0.47,P(B)=0.37 , and P(A and B)=0.15 , draw a Venn diagram, fill in the probabilities associated with the various regions, and thus determine:
a) P(A or B) b) P(A' and B') c) P(A/B)
In a random sample of 60 refrigerators, the mean repair cost was $150 and the standard deviation was $15.50. Construct a 90% confidence interval for the population mean repair cost.
(I'd like to get a second opinion on this, as I have just started statistics, and I had never seen a Venn diagram before last month. I'll ask my stats teacher tomorrow when I'm in if you like, or maybe a more advanced member can tell me if I've got this right).
P(A or B) means the chances that A or B occur. So add the A circle and the B circle.. Remember that the 0.15 can occur both in A and B, so make sure you count it, but be careful not to count it twice. (I work it out to be 0.99)
P(A' and B') means the complement of A (the total of A not occurring) and the Compliment of B (the total of B not occurring). I'm hoping someone h=will correct me if I'm wrong, but I'd read that as A not occurring and B not occurring. When you add up all the decimals, you get 0.99. That means that the chances of none of these events happening is 0.01) as the total must be 1).
P(A/B) means given A, The chances of B then happening. So if A has happened (this is the denominator) what is the chances of B happening (the numerator).
I really need to check my workings, because I am unsure whether the intersection has been counted twice in my workings, I believe it has. But the notation is right: The A' means compliment (anything that is not A) and A/B means given A happening, chances of B happening.
Don't think outside the box. Think there is no box
Did you first get the probability of .05 from a table?
Where n is the sample size, m is the mean, s is the standard deviation.
Can you do the rest?
I am getting (146.7082,153.2917)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
u also need to look up the appropriate z value for a 90% confidence interval, which is, 1.65 for 90%. These z values are available from a number of resources.