Question

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.?

Short answer

b) :

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.

c) : 50% of cups dispensed by the machine contains less than 6 fi oz.

d) : 40% of cups dispensed by the machine contains is between 5.9 and 6.1 fi oz

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

Step-by-step solution

Step 1. Given 

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.

Step 2. Part ( a ) 

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

Step 3. Part ( b ) 

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.

Step 4. Part ( c ) 

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.

Step 5. Part ( d ) 

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

Step 6. Part ( e ) 

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.