A soft drink company fills two-liter bottles on several different lines of production equipment. The fill volumes are normally distributed with a mean of 1.97 liters and a variance of 0.04 (liter)2 (i.e. a standard deviation of 0.2 liter) a. Find the probability that a randomly selected two-liter bottle would contain between 1.95 and 2.03 liters. b. If X is the fill volume of a randomly selected two-liter bottle, find the value of x for which P(X > x) = 0.3228.

Answer:a) There is a 15.62% probability that a randomly selected two-liter bottle would contain between 1.95 and 2.03 liters.b) The score is [tex]X = 2.062[/tex]Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by[tex]Z = \frac{X - \mu}{\sigma}[/tex]After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.In this problem, we have thatThe fill volumes are normally distributed with a mean of 1.97 liters and a variance of 0.04 (liter)2 (i.e. a standard deviation of 0.2 liter). This means that [tex]\mu = 1.97, \sigma = 0.2[/tex].a. Find the probability that a randomly selected two-liter bottle would contain between 1.95 and 2.03 liters.This is the pvalue of the Z score of X = 2.03 subtracted by the pvalue of the Z score of X = 1.95.X = 2.03[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{2.03 - 1.97}{0.2}[/tex][tex]Z = 0.3[/tex][tex]Z = 0.3[/tex] has a pvalue of 0.6179X = 1.95[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{1.95 - 1.97}{0.2}[/tex][tex]Z = -0.1[/tex][tex]Z = -0.1[/tex] has a pvalue of 0.4617.This means that there is a 0.6179-0.4617 = 0.1562 = 15.62% probability that a randomly selected two-liter bottle would contain between 1.95 and 2.03 liters.b. If X is the fill volume of a randomly selected two-liter bottle, find the value of x for which P(X > x) = 0.3228.This is the value of X when Z has a pvalue of 1-0.3228 = 0.6772. This is when [tex]Z = 0.46[/tex]. So:[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]0.46 = \frac{X - 1.97}{0.2}[/tex][tex]X - 1.97 = 0.46*0.2[/tex][tex]X = 2.062[/tex]The score is [tex]X = 2.062[/tex]