Question

A pharmaceutical manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The force in kilograms (kg) applied to the tablets varies a bit, with the N(11.4, 0.3) distribution. The process specifications call for applying a force between 11.3 and 12.3 kg. (a) What percent of tablets are subject to a force that meets the specifications? % (b) The manufacturer adjusts the process so that the mean force is at the center of the specifications, μ = 11.8 kg. The standard deviation remains 0.3 kg. What percent now meet the specifications? %

Answer 1

SOLUTION:

From given data,

A pharmaceutical manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The force in kilograms (kg) applied to the tablets varies a bit, with the N(11.4, 0.3) distribution. The process specifications call for applying a force between 11.3 and 12.3 kg.

N(11.4, 0.3) distribution

Mean = = 11.4

Standard deviation = = 0.3

applying a force between 11.3 and 12.3 kg.

z = ( X - ) /

(a) What percent of tablets are subject to a force that meets the specifications

at X = 11.3

z = ( 11.3 - 11.4 ) / 0.3

= -0.1 /0.3

= -0.33

at X = 12.3

z = ( 12.3 - 11.4 ) / 0.3

= 0.9 /0.3

= 3

P( 11.3 < X < 12.3 ) =  P( -0.33  < z  < 3)

= P( Z < 3) - P( Z < -0.33 )

= 0.9987 - 0.3707

= 0.628

= 62.8%

(b) The manufacturer adjusts the process so that the mean force is at the center of the specifications, μ = 11.8 kg. The standard deviation remains 0.3 kg. What percent now meet the specifications

Mean = = 11.8

Standard deviation = = 0.3

at X = 11.3

z = ( 11.3 - 11.8 ) / 0.3

= -0.5 /0.3

= -1.66

at X = 12.3

z = ( 12.3 - 11.8 ) / 0.3

= 0.5 /0.3

= 1.66

P( 11.3 < X < 12.3 ) =  P( -1.66  < z  < 1.66)

= P( Z < 1.66) - P( Z < -1.66 )

= 0.9515 - 0.0485

= 0.903

= 90.3%