Respuesta :
Answer:
|Z| < 2, which means that it would not be unusual for the mean of a sample of 3 to be 115 or more.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
If [tex]|Z| > 2[/tex], the value of X is considered to be unusual.
Central Limit Theorem
The Central Limit Theorem establishes 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.
Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15.
This means that [tex]\mu = 100, \sigma = 15[/tex]
Sample of 3
This means that [tex]n = 3, s = \frac{15}{\sqrt{3}}[/tex]
Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?
We have to find the z-score.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{115 - 100}{\frac{15}{\sqrt{3}}}[/tex]
[tex]Z = 1.73[/tex]
|Z| < 2, which means that it would not be unusual for the mean of a sample of 3 to be 115 or more.
