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Answer:
For all samples of size 5, the probability that the sample mean will be greater than 22 pounds is approximately 0.05.
Step-by-step explanation:
To solve this question, the normal probability distribution and the central limit theorem were used.
Normal probability distribution
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.
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.
For samples of size 5, which of the following is the best interpretation of Plä > 22) = 0.05
No matter the size of the sample, the mean will be the same.
However, the standard deviation will be changed.
P(ä > 22) is the probability of the sample means being above 22.
So the correct interpretation is:
For all samples of size 5, the probability that the sample mean will be greater than 22 pounds is approximately 0.05.
The probability for all sample of 5 that the sample mean will be greater than 22 pounds is approximately 0.05.
Given that ;
The distribution of backpack with mean deviation = 19.7 pounds
And the standard deviation = 3.1 pounds
According to this question,
He normal probability distribution and the central limit theorem were used.
- Normal probability distribution, ; Problems of normally distributed samples are solved using the z-score formula.
Mean standard deviation , the z-score of a measure X is given by:
Z = [tex]\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 p value,
We get the value probability that of the measure is greater than X.
- Central Limit Theorem; The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For samples of size 5, which of the following is the best interpretation of P(ä > 22) = 0.05
No matter the size of the sample mean will be the same.
However, the standard deviation will be changed.
P(ä > 22) is the probability of the sample means being above 22.
For all samples of size 5, the probability that the sample mean will be greater than 22 pounds is approximately 0.05.
For more information about standard mean and deviation click the link given below.
https://brainly.in/question/33655421
