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

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 $X = 2.062$

Step-by-step 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

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2.03 - 1.97}{0.2}$

$Z = 0.3$

$Z = 0.3$ has a pvalue of 0.6179

X = 1.95

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.95 - 1.97}{0.2}$

$Z = -0.1$

$Z = -0.1$ 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:

$Z = \frac{X - \mu}{\sigma}$

$0.46 = \frac{X - 1.97}{0.2}$

$X - 1.97 = 0.46*0.2$

$X = 2.062$

The score is $X = 2.062$

Answer 2

Answer:

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

Step-by-step 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.