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Natasha2012 [34]
2 years ago
5

Marie will earn $15 per hour at a new job. During training, she will earn $10 per hour. What percent of Marie’s regular hourly r

ate will she earn during training?
Mathematics
1 answer:
Hitman42 [59]2 years ago
4 0

Answer: 66.67%

Step-by-step explanation:

Given: The amount Marie will earn at a new job = $15 per hour

The amount she will earn during the training = $10 per hour

The percent of Marie’s regular hourly rate she will earn during training is given by :-

P=\frac{\text{Rate of amount she earn in training }}{\text{Rate of amount she earn at new job }}\times100\\\\\Rightarrow\ P=\frac{10}{15}\times100\\\\\Rightarrow\ P=66.66666\%\approx66.67\%

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0.8665 = 86.65% probability that the sample mean would be at least $39000

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:

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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.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

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Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

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2 years ago
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