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lys-0071 [83]
3 years ago
15

It took Michael 279 hours to drive home because of traffic. His sister drove the same distance in 12 of Michael’s time. How long

did the drive take Michael’s sister?
Mathematics
1 answer:
NemiM [27]3 years ago
6 0

23.25 hours because if you can divide 279 hours (michael) divided by  12, you would get 23 hours and 25minutes

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An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th
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Answer:

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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

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.

Mean of 17.42 ppm and a standard deviation of 3.25 ppm.

This means that \mu = 17.42, \sigma = 3.25

Sample of 12:

This means that n = 12, s = \frac{3.25}{\sqrt{12}}

Find the probability that the mean printing speed of the sample is greater than 18.12 ppm.

This is 1 subtracted by the p-value of Z when X = 18.12.

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

By the Central Limit Theorem

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

Z = \frac{18.12 - 17.42}{\frac{3.25}{\sqrt{12}}}

Z = 0.75

Z = 0.75 has a pvalue of 0.773.

1 - 0.773 = 0.227

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

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