Respuesta :
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 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 [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 0.6179
X = 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]
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.
