Respuesta :
Answer:
0.281 = 28.1% probability that the mean is less than 50.2 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 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.
Mean 51 and standard deviation 9 inches.
This means that [tex]\mu = 51, \sigma = 9[/tex]
A sample, with size n = 43
By the Central Limit Theorem, [tex]s = \frac{9}{\sqrt{43}} = 1.3725[/tex]
What is the probability that the mean is less than 50.2 inches?
This is the pvalue of Z when X = 50.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{50.2 - 51}{1.3725}[/tex]
[tex]Z = -0.58[/tex]
[tex]Z = -0.58[/tex] has a pvalue of 0.281
0.281 = 28.1% probability that the mean is less than 50.2 inches.
