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Answer :
Answer:
0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 16.15, \sigma = 0.45, n = 35, s = \frac{0.45}{\sqrt{35}} = 0.0761[/tex]
What is the probability that the mean of this sample is less than 15.99 ounces of water?
This is the pvalue of Z when X = 15.99. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15.99 - 16.15}{0.0761}[/tex]
[tex]Z = -2.1[/tex]
[tex]Z = -2.1[/tex] has a pvalue of 0.0179
0.0179 = 1.79% probability that the mean of this sample is less than 15.99 ounces of water.
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Rewritten by : Barada
Answer:
0.0177
Step-by-step explanation:
New SD = .45/SQRT(35) = 0.076064
P(x < 15.99), in Excel
=NORM.DIST(15.99,16.15,0.076064,TRUE)
