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# Probability of a mean of a discrete random variable distribution watch

1. Hey,
So I'm no stats expert (Studying C3 and S1 at the moment) but I have come across a problem which seems to have confused even my AS Statistics friends. It goes as so:

Code:
Marian belongs to the Handchester Building Society. She frequently visits her local branch to pay instalments on her mortgage. The number of people queuing to be served when she enters the branch may be modelled by the random variable X with the following Probability Distribution:
(format x|P(x=X))
0|0.12
1|0.33
2|0.27
3|0.18
4|0.07
5|0.03

Find, approximatly, the probability that in 100 visits to the branch, the mean number of people queuing to be served when she enters is two or more.
Now i know the solution to be 0.0967 but the workings to get here is what I'm confused about.
I've already calculated the mean, expected values and standard deviation but this question has confused me and everyone I've asked. If someone could head me in the right direction that would be great,
Thanks,
Jack.
2. You've calculated the mean, variance etc. so I won't go into details on that. Use a normal approximation to as n>30.(central limit theorem)

From table P(Z<z) = 0.90320
So the answer is 1-P(Z<z) = 0.0968
3. (Original post by jarthur36)
Hey,
So I'm no stats expert (Studying C3 and S1 at the moment) but I have come across a problem which seems to have confused even my AS Statistics friends. It goes as so:

Code:
Marian belongs to the Handchester Building Society. She frequently visits her local branch to pay instalments on her mortgage. The number of people queuing to be served when she enters the branch may be modelled by the random variable X with the following Probability Distribution:
(format x|P(x=X))
0|0.12
1|0.33
2|0.27
3|0.18
4|0.07
5|0.03

Find, approximatly, the probability that in 100 visits to the branch, the mean number of people queuing to be served when she enters is two or more.
Now i know the solution to be 0.0967 but the workings to get here is what I'm confused about.
I've already calculated the mean, expected values and standard deviation but this question has confused me and everyone I've asked. If someone could head me in the right direction that would be great,
Thanks,
Jack.

(Original post by morgan8002)
You've calculated the mean, variance etc. so I won't go into details on that. Use a normal approximation to as n>30.(central limit theorem)

From table P(Z<z) = 0.90320
So the answer is 1-P(Z<z) = 0.0968
morgan8002 has modelled this correctly, however the likelyhood for this question to appear in any exam/ in any board is very low as it combines discrete random variables with sampling distribution of the mean which do not appear together in no board.

A very good question if stats is your thing!

4. (Original post by morgan8002)
You've calculated the mean, variance etc. so I won't go into details on that. Use a normal approximation to as n>30.(central limit theorem)

From table P(Z<z) = 0.90320
So the answer is 1-P(Z<z) = 0.0968
Thanks alot for the comprehensive answer.

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