A machine used to regulate the amount of a certain chemical dispensed in the production of a particular type of cough syrup can be set so that it discharges an average of milliliters (ml) of the chemical in each bottle of cough syrup. The amount of chemical placed into each bottle of cough syrup is known to have a normal distribution with a standard deviation of 0.250 ml. If this machine discharges more than 2 ml of the chemical when preparing a given bottle of this cough syrup, the bottle is considered to be unacceptable by industry standards. Determine the setting for so that no more than 1% of the bottles of cough syrup prepared by this machine will be rejected.

The mean value of the chemical to be added to every bottle of syrup so that the bottle wont be rejected would be = 1.41750 ml

Step-by-step explanation:

Standard deviation ( σ )= 0.250 ml

assume the required amount of chemical to be deposited in each syrup bottle to be X

Hence the distribution of X in each bottle can be represented as :

X Г N ( u , 0.250 )

for a bottle not to be rejected  :  P ( X ≤ 2 )

To determine the setting so that no more than 1% of the bottles of cough syrup prepared by this machine will be rejected we have to standardize the variable X

at X = 2 ml

z = = z ----------- ( 1 )

probability for a bottle to be rejected : P( x >2 ) = 1 - P ( x ≤ 2 )

                                                                             = 0.01

attached below is a normal distribution curve showing probability values

from the curve

p( 0 < Z < z ) = 0.5 - 0.01 = 0.49  

Area = 0.49 hence from the Normal table where area = 0.49 the value of z = 2.33

Referring back to equation 1 above:  z = = 2.33 --------- (2)

u = mean value that should be added so that the bottle will not be rejected

make u subject of the equation : u = 2 - ( 2.33 * 0.250 )

                                                       u = 1.41750

The mean value of the chemical to be added to every bottle of syrup so that the bottle wont be rejected would be = 1.41750 ml

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3