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brilliants [131]
3 years ago
6

A student is running a 5-kilometer race. He runs 1 kilometer every 3 minutes. Select the function that describes his distance fr

om the finish line after x minutes.
Mathematics
1 answer:
Rom4ik [11]3 years ago
5 0
Alright, so we know that the race is 5 kilometers, so the equation will be 5-<some value>= <distance from finish line>. We also know that the student runs a kilometer every three minutes, so 3x=1km . Multiplying both sides by 5, we get 15y=5km (y being the number to make the equation make sense, or the slope). When the student has run 5km, the distance from the student to the finish line should be 0, so we get that 5-5=0, and plugging 15y in for 5 we get 5-15y=0. For 15x to equal 5, 3y=1 and y=1/3. Therefore, we plug that in for y, getting 5-15(1/3)=0. However, we have to make it for all times! Since 15 represents the minutes, we make that x, and since 0 represents the distance remaining, we make that the distance remaining, making it 5-(1/3)x=distance left. You can also think of y as the slope in y=mx+b - it stays constant that way.



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The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

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:

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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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Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

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Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

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Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

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So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

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