# Math Help - elevator weight question! urgent help !

1. ## 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

2. Originally Posted by tofustar
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 $\Pr( Y < 1120)$.

3. Originally Posted by mr fantastic
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 $\sigma^2= 8^2 \times 16$

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

4. Originally Posted by CaptainBlack
$\sigma=8 \sqrt{16}$

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: $X = W_1 + W_2 + .... + W_n \Rightarrow Var (X) = n Var (W) \Rightarrow sd (X) = \sqrt{n} sd (W)$