Question

Put a lid on it! The supplier is considering two changes to reduce to 1% the

percentage of its large-cup lids that are too small. One strategy is to adjust the

mean diameter of its lids. Another option is to alter the production process,

thereby decreasing the standard deviation of the lid diameters,

a. If the standard deviation remains at o = 0.02 inch, at what value should the

supplier set the mean diameter of its large-cup lids so that only 1% are too

small to fit?

b. If the mean diameter stays at = 3.98 inches, what value of the standard

deviation will result in only 1% of lids that are too small to fit?

c. Which of the two options in parts (a) and (b) do you think is preferable? Justify

your answer. (Be sure to consider the effect of these changes on the percent of

lids that are too large to fit.)

Answer 1

Answer:

A) 3.996 inches

B) 0.01

C) option B

Step-by-step explanation:

Note: The diameter of the lid is between : 3.95 and  4.05

A) calculate the supplier set mean diameter

std = 0.02 inch

P( x < 3.95 ) = 1% = 0.01

= P ( Z < $(\frac{3.95 - u }{0.02})$ ) = P ( Z < -2.3263 ) = 0.01

therefore :

$(\frac{3.95 - u }{0.02})$  = -2.3263

hence : u = 3.996 inches ( mean diameter )

B) At mean diameter = 3.98 calculate the value of std

P ( X < 3.95 ) = 0.01

= P ( Z < $( \frac{3.95-3.98 }{std } )$ )  = P ( Z < -2.3263 ) = 0.01

therefore

$( \frac{3.95-3.98 }{std } )$ = -2.3263

hence std = 0.01 inch

C) option B is preferable because its mean diameter is smaller and  the percent of lids too large to fit is considered more carefully using option B