Q. 24 - Elementary Statistics

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.?

Step-by-Step Solution

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 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.