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tia_tia [17]
3 years ago
6

Suppose the gpa scores of the students in a class are normally distributed with a mean of 3.4 and a standard deviation of 0.2 .

Use the empirical rule to find the percentage of students in this class with a GPA
Between 3.2 and 3.6 ( 1 standard deviation)
Mathematics
1 answer:
hoa [83]3 years ago
6 0

Suppose the gpa scores of the students in a class are normally distributed with a mean of 3.4 and a standard deviation of 0.2

We need to find how far 3.2 from 3.4

3.4 - 3.2 = 0.2  that is 1 standard deviation from mean (left)

Also we find how far 3.6 from 3.4

3.6 - 3.4 = 0.2  that is 1 standard deviation from mean (right)

As per empirical rule , 68% of values fall within 1 standard deviation from mean

so 68% of students in this class with a GPA  Between 3.2 and 3.6


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Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

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 \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

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\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

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

By the Central Limit Theorem

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

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

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