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Old September 17th, 2009, 01:55 PM
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Default elevator weight question! urgent help !

normal distribution... the max elevator weight cannot be greater than 1120kg or 16 people. suppose that the weights of lift users are normally distributed with a mean of 68kg and a standard deviation with 8kg.

what is the probability that 16 people will exceed the weight limit of 1120kg?

and also what is the probability that 18 people will not exceed the weight limit?


p.s thank you for your help
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Old September 18th, 2009, 07:05 AM
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normal distribution... the max elevator weight cannot be greater than 1120kg or 16 people. suppose that the weights of lift users are normally distributed with a mean of 68kg and a standard deviation with 8kg.

what is the probability that 16 people will exceed the weight limit of 1120kg?

and also what is the probability that 18 people will not exceed the weight limit?


p.s thank you for your help
Weight of 16 people: X ~ Normal \left( \mu = 16 \times 68 = ...., \sigma = 16 \times 8 = ....\right).

Calculate \Pr( X > 1120).


Weight of 18 people: Y ~ Normal \left( \mu = 18 \times 68 = ...., \sigma = 18 \times 8 = .... \right).

Calculate \Pr( Y < 1120).
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Old September 18th, 2009, 09:12 AM
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Weight of 16 people: X ~ Normal \left( \mu = 16 \times 68 = ...., \sigma = 16 \times 8 = ....\right).

Calculate \Pr( X > 1120).
\sigma=8 \sqrt{16}

or \sigma^2= 8^2 \times 16


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Weight of 18 people: Y ~ Normal \left( \mu = 18 \times 68 = ...., \sigma = 18 \times 8 = .... \right).

Calculate \Pr( Y < 1120).
same again but with 18 in place of 16.

CB
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Last edited by CaptainBlack; September 18th, 2009 at 09:50 AM.
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Old September 18th, 2009, 10:24 AM
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\sigma=8 \sqrt{16}

or \sigma^2= 8^2 \times 16




same again but with 18 in place of 16.

CB
Thanks for that catch. Careless of me. (Perhaps my evil nature asserting itself).

@OP: X = W_1 + W_2 + .... + W_n \Rightarrow Var (X) = n Var (W) \Rightarrow sd (X) = \sqrt{n} sd (W)
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Last edited by mr fantastic; September 18th, 2009 at 10:39 AM.
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