Question

Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people squeeze into an elevator designed to hold 4300 pounds.
a)What is the probability that the total weight exceeds the design limit
b)What design limit is exceeded by 25 occupants with probability .0001?

Answer 1

Answer:

a) $\bf 0.3446^{25}=2.7095*10^{-12}$

b) 3,624.25 pounds

Step-by-step explanation:

a)

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

In Excel and OpenOffice Calc, this value is found with the formula

=1-NORMDIST(172;160;30;1)

(NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172)

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

$\bf 0.3446^{25}=2.7095*10^{-12}$

b)

Similarly, we must find a weight w such that if p is the area to the left of w, then  

$\bf p^{25}>0.0001$  

so  

$\bf p=\sqrt[25]{0.0001}=0.6918$

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

In Excel and OpenOffice Calc this is found with

=NORMINV(1-0.6918;160;30)

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds