Dispensing Coffee. A coffee machine is supposed to dispense 6 fluid ounces (fi oz) of coffee into a paper cup. In reality, the amounts dispensed vary from cup to cup. In fact, the amount dispensed. in fi oz. is a variable with density curve y=2 for 5.75< x < 6.25, and y = 0 otherwise.

a. Graph the density curve of this variable.

b. Show that the area under this density curve to the left of any number x between 5.75 and 6 .25 equals 2x-11.5. What percentage of cups dispensed by this machine contain

c. less than 6 fi oz?

d. between 5.9 and 6.1 11 oz

e. at least 5.8 fi oz.?

A coffee machine is supposed to dispense 6 fluid ounces (fi oz) of coffee into a paper cup , with density curve y=2 for 5.75< x < 6.25, and y = 0 otherwise.

Graph the density of the given variable.

In the given situation, the random variable (x) is defined as the amount dispensed, in fi oz The density function is given by, y 2,5.75<x<6.25 and y =0 otherwise.

The graph of the density function is given below

Show that the area under the density curve to the left of any number x between 5.75 and 6.255

equal to 2x-11.5

For the density function in part (8), base of the rectangle is (x-5.75) and height of the

rectangle is 2 Thus, the area of the rectangle is,

Area = (Base) (Height)

= (x-5.75) (2)

=2x-11.5

Hence, the area under the density curve to the left of any number x between 5 .75 and 6.25 is

equal to 2x-11.5.

Find the percentage of cups dispensed by the machine contains less than 6 fi oz

(Area to the less than 6 fl oz) = 2(6)-11.5 [From part (b),x=6]

=12-11.5 =0.5

Thus 50% of cups dispensed by the machine contains less than 6 fi oz.

Find the percentage of cups dispersed by the machine contains is between 5.9 and 6.1 oz

(Area between 5.9 and 6.1) = (Area to the left of 6.1)-(Area to the left of 5.9)

=[(2x6.1)-11.5]-[(2x5.9)-11.5] [From part (b)]

=(12.2-11.5)-(11.8-11.5)

= 0.7-0.3 = 0.4

Thus 40% of cups dispensed by the machine contains is between 5.9 and 6.1 oz

Find the percentage of cups dispensed by the machine contains at least than 5,8 fi oz

(Area to the right of 5.8) = 1-(Area to the left of 5.8)

=1-[2(5.8)-11.5] [From part: (b)]

=1-(11.6-11.5)

=1-0.1

=0.9

Thus 90% of cups dispensed by the machine contains at least than 5.8 fi oz.