Mathematics, 06.04.2020 18:19 yovann

The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a standard deviation of .29.Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

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Answer from: kaymillsaps

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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:

$\mu = 3.02, \sigma = 0.29, n = 20, s = \frac{0.29}{\sqrt{20}} = 0.0648$

Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

This is 1 subtracted by the palue of Z when X = 3.10. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.1 - 3.02}{0.0648}$

Z = 1.23

Z = 1.23 has a pvalue of 0.8907

1 - 0.8907 = 0.1093

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

Answer from: Quest

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

Arectangle is transformed according to the rule r0,90 the image of the rectangle has vertices locate

Answer from: Quest

Its d so thats how it would would back from 1.0 form three four two

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The GPAs of all students enrolled at a large university have an approximately normal distribution wi...