SAT verbal scores are normally distributed with a mean of 430 and a standard deviation of 120 (based on data from the College Board ATP). (a) If a single student is randomly selected, find the probability that the sample mean is above 500. (b) If a sample of 35 students are selected randomly, find the probability that the sample mean is above 500. These two problems appear to be very similar. Which problem requires the application of the central limit theorem, and in what way does the solution process differ between the two problems?

Answers

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a:

a) 0.281 = 28.1% probability that the sample mean is above 500.

b) 0.0003 = 0.03% probability that the sample mean is above 500.

In a normal distribution 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]

It 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, which is the percentile of X. By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

The mean is of 430, hence [tex]\mu = 430[/tex].The standard deviation is of 120, hence [tex]\sigma = 120[/tex].

Item a:

The probability is the p-value of Z when X = 500, hence:

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

[tex]Z = \frac{500 - 430}{120}[/tex]

[tex]Z = 0.58[/tex]

[tex]Z = 0.58[/tex] has a p-value of 0.719.

1 - 0.719 = 0.281

0.281 = 28.1% probability that the sample mean is above 500.

Item b:

Sample of 35, hence [tex]n = 35, s = \frac{120}{\sqrt{35}}[/tex]

Then:

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

By the Central Limit Theorem

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

[tex]Z = \frac{500 - 430}{\frac{120}{\sqrt{35}}}[/tex]

[tex]Z = 3.45[/tex]

[tex]Z = 3.45[/tex] has a p-value of 0.9997.

1 - 0.9997 = 0.0003

0.0003 = 0.03% probability that the sample mean is above 500.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213


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rgb

Step-by-step explanation:

wrtbtrb

a) The null hypothesis is: p = 0.8 and the alternative hypothesis is p ≠ 0.8.

b) p-value is 0.0838

c) We accept the null hypothesis

What are null and alternative hypothesis?

The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

What is p-value?

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The alternative hypothesis is

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