Asoft 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.

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

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by

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 that

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). This means that .

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

has a pvalue of 0.6179

X = 1.95

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 . So:

The score is

$97520

step-by-step explanation:

add 6% to the value she got from selling the house.