Respuesta :
Answer:
15.25% probability that the bottle contains between 12.06 and 12.12 ounces.
Step-by-step explanation:
Problems of normally distributed samples are 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]
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 12.16, \sigma = 0.04[/tex]
Find the probability that the bottle contains between 12.06 and 12.12 ounces.
This is the pvalue of Z when X = 12.12 subtracted by the pvalue of Z when X = 12.06. So
X = 12.12
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.12 - 12.16}{0.04}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
X = 12.06
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12.06 - 12.16}{0.04}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062
0.1587 - 0.0062 = 0.1525
15.25% probability that the bottle contains between 12.06 and 12.12 ounces.
The probability that the bottle contains between 12.06 and 12.12 ounces is 15.25%
Z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation\\\\\\Given\ that\ \mu=12.16,\sigma=0.04\\\\\\a) For\ x=12.06:\\\\z=\frac{12.06-12.16}{0.04}=-2.5\\\\For \ x=12.12:\\\\z=\frac{12.12-12.16}{0.04}=-1[/tex]
From the normal distribution table, P(12.06 < x < 12.12) = P(-2.5 < z < -1) = P(z < -1) - P(z < -2.5) = 0.1587 - 0.0062 = 15.25%
The probability that the bottle contains between 12.06 and 12.12 ounces is 15.25%
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