Atype of elevator has a maximum weight capacity y1 , which is normally distributed with mean 5000 pounds and standard deviation 300 pounds. for a certain building equipped with this type of elevator, the elevator’s load, y2, is a normally distributed random variable with mean 4000 pounds and standard deviation 400 pounds. for any given time that the elevator is in use, find the probability that it will be overloaded, assuming that y1 and y2 are independent.

 then value is 0.0228

Step-by-step explanation:

Maximum weight capacity= Y1

Elevator's loaf = Y2

probability that is overloaded :

P(Y2 > Y1)

TakingP( Y2 - Y1) > 0

E[Y1 - Y2 ] = 500 - 4000 = 1000

P( Y2 > Y1) = P( Y2 - Y1 > 0 ) = P( Y1 - Y2 < 0 ) ;

E[Y1 - Y2] = 500-4000 = 1000

P (Y2 > Y1 )

then

P (Y1 - Y2 < 0)

P((Y1-Y2)-1000)/50 = (0-1000/500)

P((Y1 - Y2 )  - 2)

 then value is 0.0228

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p