In Exercises 5–20, find the range, variance, and standard devi...

Question

In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Jaws 3 Listed below are the number of unprovoked shark attacks worldwide for the last several years. What extremely important characteristic of the data is not considered when finding the measures of variation?

70 54 68 82 79 83 76 73 98 81

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says listed below are the reaction times in seconds of 5 individuals tested in a driving simulation study. Find the range, variance and standard deviation for the given sample data. Include appropriate units such as seconds in your results. One extremely important characteristic of the data is not considered when finding the measures of variation, and we're given the data of 0.6, 0.8, 1.0, 0.9, and 0.7. We're also given four possible choices as our answers. For choice A, we have the range is equal to 0.8 seconds. The variance is equal to 0.034 seconds squared. The standard deviation is equal to 0.28 seconds, and the measures of variation do not indicate whether the data values are evenly spread out or clustered around the mean. For choice B, we have the range is equal to 0.4 seconds. The variance is equal to 0.025 seconds squared. The standard deviation is equal to 0.16 seconds, and the measures of variation do not consider the distribution shape or potential outliers, which might impact the interpretation of reaction times. For choice C, we have the range is equal to 0.4 seconds. The variance is equal to 0.025 seconds squared. The standard deviation is equal to 0.28 seconds. And the measures of variation do not indicate the reliability of the reaction time measurements or external factors influencing them, such as fatigue or distractions. And for choice D, we have the range as equal to 0.8 seconds, the variance is equal to 0.034 seconds squared, the standard deviation is equal to 0.40 seconds, and the measures of variation do not account for the increasing trend of reaction times over multiple trials, which could indicate learning effects. Now, the first thing we need to do to solve this problem is to find the range, variance and standard deviation of our data set. So we're gonna start off by finding the range. And recall for the range. That that is going to be equal to the maximum value minus the minimum value, which in this case is going to be equal to 1.0, minus 0.6, which is equal to 0.4 seconds. And so that is the first quantity that we needed to find. Now, for the next quantity, we're going to be looking for the variance Squad. And we call your formula for the variant, that's going to be equal to the summation of the quantity of X of I minus X in quantity squared, divided by the quantity of N minus 1 in quantity. Where X of I is an individual data point, X bar is our mean, and N is the number of data points, which in this case is 5. Now, in order to use this formula, we'll first need to calculate our mean X bar. And recall to calculate the mean that we just need to add up all of our data points and divide by the number of data points. So X bar is going to be equal to the quantity of 0.6 plus 0.8 plus 1.0 plus 0.9 plus 0.7 in quantity, and all of that is going to be divided by the number of data points that we have, which is 5. And evaluating this this expression, that means that X bar is going to be equal to 0.8 seconds. So now we can actually calculate our variance as squared. And so using our variance formula, S2 is going to be equal to, and the numerator will have the quantity of 0.6 minus 0.8 in quantity squared. Plus, the quantity of 0.8 minus 0.8 in quantity squared. Plus, the quantity of 1.0 minus 0.8 in quantity. Where? Plus quantity of 0.9 minus 0.8 in quantity squared. Plus, the quantity of 0.7 minus 0.8, in quantity squared, and all of that is divided by the quantity of N minus 1, which in this case is going to be 5 minus 1. So simplifying this expression, this is going to be equal to, and the numerator for the first term, we'll have 0.6 minus 0.8, which will give us minus 0.2, and when I square that, I'll get 0.04. Then we'll have less. For the second term, we'll have 0.8 minus 0.8, which will give me 0, and squaring that will also give me 0. Then we'll have plus For the next term, we'll have 1.0 minus 0.8, which will give me 0.2, and squaring that will give me 0.04. Then we'll have plus For the next term, we'll have 0.9 minus 0.8, which is 0.1, and squaring that will give me 0.01, and then we'll have plus. For the last term we'll have 0.7 minus 0.8, which will give you minus 0.1, and squaring that will give me 0.01. And all of that is divided by 20 of 5 minus 1, which is 4. And so, evaluating this expression, that means that Squared is going to be equal to. 0 point. 0 25, and that has units of seconds squared. Since both our mean and our individual points are uh have units of seconds, and then we square that in the numerator, which will give us units of seconds squared. So this here is the second quantity that we need to find. For the last quantity, we need to find the standard deviation S. That's just equal to the square root of our variance Squad. So, this is going to be equal to the square root. of 0.025. And evaluating that expression, this is going to be approximately equal to 0.16 seconds. So here we just round it to two decimal places, and the units here are going to be seconds. Since when I take the square root of the variants, I have to take the square root of the value, but also the units. I'm going to take the square root of second square roots, I get units of seconds. So this here is the third quantity that we need to find. Now we have found all three of these quantities, we just need to answer the second half of this problem. Which is what extremely important characteristic of the data is not considered when finding the measures of variation. So, the quantities that we have calculated, the range, the variance and standard deviation, they only describe how spread out the data values are, but they do not take into account several things. First, they don't take into account the distribution shape. Measures of variation do not tell us whether the data is normally distributed, skewed, or has multiple peaks. So, for example, two data sets with the same standard deviation might have entirely different distributions. The other quantity that it doesn't tell us any information about is outliers. And extreme values such as outliers can heavily influence the range and variants, but the numerical measures alone do not explicitly highlight their presence. So, for example, if we had a single reaction time with a much longer value, say of 2 seconds, that would significantly affect our calculations, specifically the range and Um, the standard deviation. And our formulas do not really differentiate between Uh, data points that have normal variability versus outliers. And the other thing that we do not take into account are contextual factors. The variation measures do not explain the underlying reasons behind reaction time differences. So, factors such as like fatigue and distractions or or any of the other conditions under which the test was conducted, that could impact reaction times, but these are not reflected in our numerical calculations. So taking all of this into account, If we look at our answer choices, We see that the correct answer choice corresponds to answer choice B. Since this answer choice had the same values that we had calculated for the range, the variance, and standard deviation, and it also had the correct explanation. Or what we have not considered with our data, which is the measures variations do not consider the distribution shape or potential alliesers which might impact the interpretation of reaction times. So, I hope that this explanation has been useful, and I'll see you in the next video.

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Key Concepts

Measures of Variation

Sample Data

Outliers

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