QUESTION IMAGE
Question
here is a data set. 21, 19, 26, 24, 12, 17, 23, 18, 26, 24, 18, 24. find the mean, median, mode, range, and standard deviation of the new data set obtained after multiplying each value by 5 in the data set above. round your answer for the standard deviation to the nearest thousandth. mean of the new data set: median of the new data set: mode of the new data set: range of the new data set: standard deviation of the new data set:
Step1: Multiply each data - point
The original data set is \(21,19,26,24,12,17,23,18,26,24,18,24\). After multiplying each value by \(5\), the new data set is \(105,95,130,120,60,85,115,90,130,120,90,120\).
Step2: Calculate the mean
The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 12\) and \(\sum_{i=1}^{12}x_{i}=105 + 95+130+120+60+85+115+90+130+120+90+120 = 1260\). So, \(\bar{x}=\frac{1260}{12}=105\).
Step3: Calculate the median
First, arrange the data in ascending order: \(60,85,90,90,95,105,115,120,120,120,130,130\). Since \(n = 12\) (even), the median is the average of the \(\frac{n}{2}\) - th and \((\frac{n}{2}+1)\) - th ordered values. \(\frac{105 + 115}{2}=110\).
Step4: Calculate the mode
The mode is the value that appears most frequently. In the data set \(60,85,90,90,95,105,115,120,120,120,130,130\), the mode is \(120\).
Step5: Calculate the range
The range is the difference between the maximum and minimum values. Range \(=130 - 60=70\).
Step6: Calculate the standard deviation
The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
\(\sum_{i=1}^{12}(x_{i}-105)^{2}=(105 - 105)^{2}+(95 - 105)^{2}+(130 - 105)^{2}+(120 - 105)^{2}+(60 - 105)^{2}+(85 - 105)^{2}+(115 - 105)^{2}+(90 - 105)^{2}+(130 - 105)^{2}+(120 - 105)^{2}+(90 - 105)^{2}+(120 - 105)^{2}\)
\(=0+100 + 625+225 + 2025+400+100+225+625+225+225+225=4800\).
\(s=\sqrt{\frac{4800}{11}}\approx20.817\).
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Mean of the new data set: \(105\)
Median of the new data set: \(110\)
Mode of the new data set: \(120\)
Range of the new data set: \(70\)
Standard deviation of the new data set: \(20.817\)