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Math  /  Data & Statistics

QuestionFifteen students were selected and asked how many hours each studied for the final exam in statistics. Their answers are recorded here. 2690239497710416\begin{array}{lllllllllllllll}2 & 6 & 9 & 0 & 2 & 3 & 9 & 4 & 9 & 7 & 7 & 10 & 4 & 1 & 6\end{array} Send data to Excel
Find the range, variance, and standard deviation. Round the variance to one decimal place and the standard deviation to two decimal places, if necessary.
The range is □\square 10 hours.
Part: 1/31 / 3
Part 2 of 3
The variance is □\square

Studdy Solution

STEP 1

1. The data set consists of 15 values representing the number of hours each student studied.
2. The range is the difference between the maximum and minimum values.
3. Variance is calculated as the average of the squared differences from the mean.
4. Standard deviation is the square root of the variance.

STEP 2

1. Calculate the range.
2. Calculate the variance.
3. Calculate the standard deviation.

STEP 3

List the data points:
2,6,9,0,2,3,9,4,9,7,7,10,4,1,6 2, 6, 9, 0, 2, 3, 9, 4, 9, 7, 7, 10, 4, 1, 6

STEP 4

Identify the minimum and maximum values in the data set:
Minimum value = 0
Maximum value = 10

STEP 5

Calculate the range:
Range=Maximum value−Minimum value=10−0=10 \text{Range} = \text{Maximum value} - \text{Minimum value} = 10 - 0 = 10

STEP 6

Calculate the mean of the data set:
Mean=2+6+9+0+2+3+9+4+9+7+7+10+4+1+615 \text{Mean} = \frac{2 + 6 + 9 + 0 + 2 + 3 + 9 + 4 + 9 + 7 + 7 + 10 + 4 + 1 + 6}{15}
Mean=7915≈5.27 \text{Mean} = \frac{79}{15} \approx 5.27

STEP 7

Calculate each squared difference from the mean, then find the average:
\begin{align*} (2 - 5.27)^2 & \approx 10.73 \\ (6 - 5.27)^2 & \approx 0.53 \\ (9 - 5.27)^2 & \approx 13.87 \\ (0 - 5.27)^2 & \approx 27.77 \\ (2 - 5.27)^2 & \approx 10.73 \\ (3 - 5.27)^2 & \approx 5.17 \\ (9 - 5.27)^2 & \approx 13.87 \\ (4 - 5.27)^2 & \approx 1.61 \\ (9 - 5.27)^2 & \approx 13.87 \\ (7 - 5.27)^2 & \approx 2.97 \\ (7 - 5.27)^2 & \approx 2.97 \\ (10 - 5.27)^2 & \approx 22.47 \\ (4 - 5.27)^2 & \approx 1.61 \\ (1 - 5.27)^2 & \approx 18.23 \\ (6 - 5.27)^2 & \approx 0.53 \\ \end{align*}

STEP 8

Sum the squared differences and divide by the number of data points to find the variance:
Variance=10.73+0.53+13.87+27.77+10.73+5.17+13.87+1.61+13.87+2.97+2.97+22.47+1.61+18.23+0.5315\text{Variance} = \frac{10.73 + 0.53 + 13.87 + 27.77 + 10.73 + 5.17 + 13.87 + 1.61 + 13.87 + 2.97 + 2.97 + 22.47 + 1.61 + 18.23 + 0.53}{15}
Variance≈146.9615≈9.8\text{Variance} \approx \frac{146.96}{15} \approx 9.8
The variance is:
9.8 \boxed{9.8}

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