Pregunta
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Find the standard deviation for the following group of data items. \( 6,11,11,19 \) The standard deviation is approximately (Round to two decimal places as needed.)

Ask by Brooks Gross. in the United States
Feb 26,2025

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The standard deviation is approximately 4.66.

Solución

To calculate the standard deviation, first determine if we're treating these data as the entire population or just a sample. Since the problem refers to “the following group of data items” without specifying that it is a sample, we will assume that these four numbers constitute the entire population. Step 1. Find the mean.   Mean = (6 + 11 + 11 + 19) / 4 = 47 / 4 = 11.75 Step 2. Find the squared difference of each data item from the mean.   For 6:  (6 − 11.75)² = (−5.75)² = 33.0625   For 11: (11 − 11.75)² = (−0.75)² = 0.5625   For 11: (11 − 11.75)² = (−0.75)² = 0.5625   For 19: (19 − 11.75)² = (7.25)² = 52.5625 Step 3. Find the variance by dividing the sum of these squared differences by the number of data items.   Sum of squared differences = 33.0625 + 0.5625 + 0.5625 + 52.5625 = 86.75   Variance = 86.75 / 4 = 21.6875 Step 4. Take the square root of the variance to obtain the standard deviation.   Standard deviation = √21.6875 ≈ 4.66 Thus, the standard deviation for the data items is approximately 4.66.

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To calculate the standard deviation, first find the mean of the data set. The mean is calculated as follows: \[ \text{Mean} = \frac{6 + 11 + 11 + 19}{4} = \frac{47}{4} = 11.75 \] Next, subtract the mean from each data item, square the result, and then find the average of those squared differences: 1. \((6 - 11.75)^2 = (-5.75)^2 = 33.0625\) 2. \((11 - 11.75)^2 = (-0.75)^2 = 0.5625\) 3. \((11 - 11.75)^2 = (-0.75)^2 = 0.5625\) 4. \((19 - 11.75)^2 = (7.25)^2 = 52.5625\) Now, calculate the variance: \[ \text{Variance} = \frac{33.0625 + 0.5625 + 0.5625 + 52.5625}{4} = \frac{86.75}{4} = 21.6875 \] Finally, take the square root of the variance to find the standard deviation: \[ \text{Standard deviation} = \sqrt{21.6875} \approx 4.65 \] Thus, the standard deviation for the group of data is approximately **4.65** (rounded to two decimal places).

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