Respuesta :
Answer:
0.9099 = 90.99% probability that their mean length is less than 21.9 inches.
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.
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.
Mean of 18.2 inches, and standard deviation of 3.9 inches.
This means that [tex]\mu = 18.2, \sigma = 3.9[/tex]
2 itens:
This means that [tex]n = 2, s = \frac{3.9}{\sqrt{2}}[/tex]
What is the probability that their mean length is less than 21.9 inches?
This is the p-value of Z when X = 21.9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{21.9 - 18.2}{\frac{3.9}{\sqrt{2}}}[/tex]
[tex]Z = 1.34[/tex]
[tex]Z = 1.34[/tex] has a p-value of 0.9099.
0.9099 = 90.99% probability that their mean length is less than 21.9 inches.
