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Marrrta [24]
3 years ago
14

The final grade in an algebra course is based on a weighted average - homework counts 10%, quizzes count 10%, tests count 50%, a

nd the final exam counts 30%. A student has a homework average of 96, quiz average of 88, and a test average of 84. What is the minimum score a student needs to score on the final exam to earn at least a 90 for the final course grade? (Round to the nearest tenth.)
Mathematics
2 answers:
erma4kov [3.2K]3 years ago
4 0

Answer:  98.7

<u>Step-by-step explanation:</u>

Homework +    Quiz    +   Test    + Final Exam = 90

  96(10%)    + 88(10%)  + 84(50%) +    x(30%)    = 90

   9.6           +    8.8      +     42      +      0.30x     = 90

                                                      60.4 + 0.30x  = 90

                                                                 0.30x = 29.6

                                                                        x = 98.7

The minimum score is 98.7

Dovator [93]3 years ago
3 0

<u>Answer:</u>

x = 98.7

<u>Step-by-step explanation:</u>

To find the minimum score a student needs to score on the final exam to earn at least a 90 for the final course grade, we will multiply the student's scores with their respective weight-age.

Supposing the score in the final exam to be x, we can write an equation:

Homework + quizzes + tests + final exam = 90

96(10%) + 88(10%) + 84(50%) + x(30%) = 90

9.6 + 8.8 + 42 + 0.3x = 90

60.4 + 0.3x = 90

0.3x = 29.6

x = 98.7

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a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

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In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

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

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

Z = \frac{205 - 200}{5}

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Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

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0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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