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Vinvika [58]
3 years ago
5

The grades on a statistics test are normally distributed with a mean of 62 and Q1=52. If the instructor wishes to assign B's or

higher to the top 30% of the students in the class, what grade is required to get a B or higher? Please round your answer to two decimal places.
Mathematics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

To get a B or higher, you need to get a grade of at least 69.77.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that

Mean of 62, so \mu = 62

Q1 of 52 means that the z score of X = 52 has a pvalue of 0.25. Z has a pvalue of 0.25 between -0.67 and -0.68, so we use Z = -0.675

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

-0.675 = \frac{52 - 62}{\sigma}

-0.675\sigma = -10

Multiplying the equality by (-1), we have that:

0.675\sigma = 10

\sigma = \frac{10}{0.675}

\sigma = 14.8

If the instructor wishes to assign B's or higher to the top 30% of the students in the class, what grade is required to get a B or higher?

Those are the Z scores that have a pvalue higher than 0.70. So we have to find X when Z has a pvalue of 0.70. This is between 0.52 and 0.53, so we use Z = 0.525

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

0.525 = \frac{X - 62}{14.8}

X - 62 = 14.8*0.525

X = 69.77

To get a B or higher, you need to get a grade of at least 69.77.

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So, Increase = New Number - Original Number. Therefore, Increase = 283 - 147 which equals 136.
136 = 283 - 147.
Then you want to divide the increase by the original number and multiply the answer by 100.
136/147 = 0.92
0.92 * 100 = 92.

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