Question

A fitness company is building a 20-story high-rise. Architects building the high-rise know that women working for the company have weights that are normally distributed with a mean of 143 lb and a standard deviation of 29 lb, and men working for the company have weights that are normally distributed with a mean of 181 lb and a standard deviation or 32 lb. You need to design an elevator that will safely carry 12 people. Assuming a worst case scenario of 12 male passengers, find the maximum total allowable weight if we want to a 0.999 probability that this maximum will not be exceeded when 12 males are randomly selected.

maximum weight = -lb

Enter your answer rounded to the nearest whole number. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

If X₁, X₂...X_n  is a sequence of n independent, identically distributed variables with

E(X_i) =

and

V(X_i) =

then a random variable which is defined as

Y = X₁ + X₂..+..X_n

will also be a normal random variable with

E(Y) = n*

V(Y) = n*

For 12 males, the weight will also be normally distributed with

mean, = 12*181

= 2172

variance = 12*(32)²

= 12288

std. deviation, = 110.85

Now in the worst case scenario (when 12 males board the lift), For it not to fail with probability 0.999 then

P(Y<y) = 0.999

We know that std. normal variable z is defined as

Z = (Y-)/

So,

P(Z<(y-)/) = 0.999

Using the normal table:

(y-)/ = 3.09023

y = 2172 + 3.09023*110.85

= 2514.55lb

~ 2515 lb

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