If the mean GPA among students is 3.25 with a standard deviation of 1.75, and we select a random sample of 300 people, at what value for the sample mean would be greater than exactly 95% of possible sample means?

Respuesta :

Answer:

[tex]X = 3.4161[/tex]

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 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.

In this problem, we have that:

[tex]\mu = 3.25, \sigma = 1.75, n = 300, s = \frac{1.75}{\sqrt{300}} = 0.101[/tex]

At what value for the sample mean would be greater than exactly 95% of possible sample means?

This is X when Z has a pvalue of 0.95. So X when Z = 1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]1.645 = \frac{X - 3.25}{0.101}[/tex]

[tex]X - 3.25 = 1.645*0.101[/tex]

[tex]X = 3.4161[/tex]