students in a science class roll a model car down a ramp. they repeat this process for 5 trials, recording…

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? use the drop - down menus to explain. model car trials\n| |trial 1|trial 2|trial 3|trial 4|trial 5|\n|--|--|--|--|--|--|\n|feet traveled|3|6|7|5|6|\nclick the arrows to choose an answer from each menu. the standard deviation of the data will choose... if the trial 1 value is removed. this is because choose...

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? use the drop - down menus to explain. model car trials\n| |trial 1|trial 2|trial 3|trial 4|trial 5|\n|--|--|--|--|--|--|\n|feet traveled|3|6|7|5|6|\nclick the arrows to choose an answer from each menu. the standard deviation of the data will choose... if the trial 1 value is removed. this is because choose...

Answer

Explanation:

Step1: Calculate original mean

The original data set is (3,6,7,5,6). The mean (\bar{x}_1=\frac{3 + 6+7+5+6}{5}=\frac{27}{5} = 5.4).

Step2: Calculate original variance

The variance (s_1^{2}=\frac{(3 - 5.4)^{2}+(6 - 5.4)^{2}+(7 - 5.4)^{2}+(5 - 5.4)^{2}+(6 - 5.4)^{2}}{5}) [ \begin{align*} &=\frac{(- 2.4)^{2}+(0.6)^{2}+(1.6)^{2}+(-0.4)^{2}+(0.6)^{2}}{5}\ &=\frac{5.76+0.36 + 2.56+0.16+0.36}{5}\ &=\frac{9.2}{5}=1.84 \end{align*} ] The original standard - deviation (s_1=\sqrt{1.84}\approx1.36).

Step3: Calculate new mean

After removing the first - trial value ((3)), the new data set is (6,7,5,6). The new mean (\bar{x}_2=\frac{6 + 7+5+6}{4}=\frac{24}{4}=6).

Step4: Calculate new variance

The new variance (s_2^{2}=\frac{(6 - 6)^{2}+(7 - 6)^{2}+(5 - 6)^{2}+(6 - 6)^{2}}{4}) [ \begin{align*} &=\frac{(0)^{2}+(1)^{2}+(-1)^{2}+(0)^{2}}{4}\ &=\frac{0 + 1+1+0}{4}\ &=\frac{2}{4}=0.5 \end{align*} ] The new standard - deviation (s_2=\sqrt{0.5}\approx0.71).

Answer:

The standard deviation of the data will decrease if the Trial 1 value is removed. This is because the value of 3 was an out - lier relative to the other values, and removing it made the data more clustered around the mean.