Answered

Students in a science class roll a model car down a ramp. They repeat this process for 5 trials, recording the number of whole feet the car travels during each trial in the table below.



The students want to count the first trial as practice and remove its value from the data set. How will the standard deviation of the data in the table change if they remove the Trial 1 value?



\begin{table}

\captionsetup{labelformat=empty}

\caption{Model Car Trials}


Trial 1Trial 2Trial 3Trial 4Trial 5

Feet Traveled		36756




\end{table}



The standard deviation of the data will [increase/decrease/stay the same] if the Trial 1 value is removed. This is because [explain].

Verified Answer :

  • The standard deviation of the initial data \(\{3, 6, 7, 5, 6\}\) is approximately \(1.356\).
  • The standard deviation of the new data \(\{6, 7, 5, 6\}\) is approximately \(0.707\).
  • The standard deviation \(\boxed{\text{decreases}}\) because \(\boxed{\text{the removed value (3) was an outlier that significantly contributed to the data's original spread}}\).

Explanation

  1. Understand the problem and provided data

We are given a data set of five trials for a model car's travel distance: \(\{3, 6, 7, 5, 6\}\). We need to determine how the population standard deviation changes when the first trial (a value of \(3\)) is removed, resulting in a new data set: \(\{6, 7, 5, 6\}\). The standard deviation measures how spread out the values in a data set are from their mean. Generally, removing a value that is far from the mean decreases the standard deviation, while removing a value near the mean increases it.

  1. Calculate the initial standard deviation

First, we calculate the statistics for the original data set \(\{3, 6, 7, 5, 6\}\).

  • The mean is \(\bar{x}_1 = \frac{3 + 6 + 7 + 5 + 6}{5} = \frac{27}{5} = 5.4\).
  • To find the variance (\(\sigma_1^2\)), we calculate the squared deviations from the mean:

\((3-5.4)^2 = (-2.4)^2 = 5.76\)
\((6-5.4)^2 = (0.6)^2 = 0.36\)
\((7-5.4)^2 = (1.6)^2 = 2.56\)
\((5-5.4)^2 = (-0.4)^2 = 0.16\)
\((6-5.4)^2 = (0.6)^2 = 0.36\)
Sum of squared deviations: \(5.76 + 0.36 + 2.56 + 0.16 + 0.36 = 9.2\).

  • The variance is \(\sigma_1^2 = \frac{9.2}{5} = 1.84\), so the standard deviation is \(\sigma_1 = \sqrt{1.84} \approx 1.356\).
  1. Calculate the new standard deviation

Next, we calculate the statistics for the new data set \(\{6, 7, 5, 6\}\) after removing the value \(3\).

  • The new mean is \(\bar{x}_2 = \frac{6 + 7 + 5 + 6}{4} = \frac{24}{4} = 6\).
  • To find the new variance (\(\sigma_2^2\)), we calculate the squared deviations from the new mean:

\((6-6)^2 = 0^2 = 0\)
\((7-6)^2 = 1^2 = 1\)
\((5-6)^2 = (-1)^2 = 1\)
\((6-6)^2 = 0^2 = 0\)
Sum of squared deviations: \(0 + 1 + 1 + 0 = 2\).

  • The new variance is \(\sigma_2^2 = \frac{2}{4} = 0.5\), so the new standard deviation is \(\sigma_2 = \sqrt{0.5} \approx 0.707\).
  1. Compare and conclude

Comparing the two values, \(\sigma_1 \approx 1.356\) and \(\sigma_2 \approx 0.707\), we see that \(\sigma_2 < \sigma_1\). The standard deviation has decreased. This is because the removed value of \(3\) was significantly further from the original mean (\(5.4\)) than the other data points were. Removing a data point that is far from the mean reduces the overall dispersion (spread) of the data.

Examples

Standard deviation is a crucial concept in analyzing variability in experimental data. For example, in a science project, if a student conducts trials to measure the distance a model car travels, they might notice that one trial is much shorter or longer than others due to a "bad start." By calculating the standard deviation before and after removing such an outlier, the student can quantify the reliability and consistency of their results. If the standard deviation drops significantly, it suggests that the removed trial was a major source of inconsistency in their measurements. This technique is widely used in fields ranging from quality control in manufacturing to analyzing performance metrics in sports.

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