Question

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 1

a) P [ 1.97 ≤ z ≤ 2.03 ] = 0.9532

b) P [ X ≥ 0.3228 ]

Note the values in the first answer were express in term of z since the attached was done with z instead of x

Explanation:

Question a:

We have to find the region between points

z1 = -1.95 and

z2 = 2.03

These two points correspond to areas:

z1 ⇒ 0.0256 and z2 ⇒ 0.9788

If you look the attached figure we realizad that our asked probability is the one between these two poins, so we have to subtract these two values and get:

P [ 1.97 ≤ z ≤ 2.03 ] = 0.9788 - 0.0256 = 0.9532

Question b:

We look in z tables for the area 0.3228 and directly find the z value of - 0.46.

Answer 2

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
`X = 2.062`

Explanation:

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

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

`Z = (X - \mu)/(\sigma)`

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
`\mu = 1.97, \sigma = 0.2`.

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

`Z = (X - \mu)/(\sigma)`

`Z = (2.03 - 1.97)/(0.2)`

`Z = 0.3`

`Z = 0.3` has a pvalue of 0.6179

X = 1.95

`Z = (X - \mu)/(\sigma)`

`Z = (1.95 - 1.97)/(0.2)`

`Z = -0.1`

`Z = -0.1` 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
`Z = 0.46`. So:

`Z = (X - \mu)/(\sigma)`

`0.46 = (X - 1.97)/(0.2)`

`X - 1.97 = 0.46*0.2`

`X = 2.062`

The score is
`X = 2.062`