Respuesta :
Answer:
97.92% probability that the mean score of your sample is between 22 and 28
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples 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 random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 25, \sigma = 6.5, n = 25, s = \frac{6.5}{\sqrt{25}} = 1.3[/tex]
What is the probability that the mean score of your sample is between 22 and 28
This is the pvalue of Z when X = 28 subtracted by the pvalue of Z when X = 22. So
X = 28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{28 - 25}{1.3}[/tex]
[tex]Z = 2.31[/tex]
[tex]Z = 2.31[/tex] has a pvalue of 0.9896
X = 22
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{22 - 25}{1.3}[/tex]
[tex]Z = -2.31[/tex]
[tex]Z = -2.31[/tex] has a pvalue of 0.0104
0.9896 - 0.0104 = 0.9792
97.92% probability that the mean score of your sample is between 22 and 28
