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It is 2.61, approx.

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9y ago
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Q: What is the standard deviation of 4 7 8 9 12?
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What is the variance and standard deviation of 4 7 2 7 9 10 15?

Variance = 17.9047619 Standard Deviation = 4.23140188


What is the standard deviation 1 4 and 7?

For 1 4 and 7: σ=3


What is the standard deviation of these four scores 2 4 3 7?

Standard deviation is the square root of the mean. The mean for this set is (2 + 4 + 3 + 7)/4 = 16/4 = 4; the square root of this is 2.


What is the standard deviation of these set of numbers 6 7 7 12 12 11 9 9 8 4 6 13 14?

For 6 7 7 12 12 11 9 9 8 4 6 13 14: σ=3.0947


What is the mean variance and standard deviation for the numbers 2 4 8 2 9 1 7 11 4 7 1 10 1 5?

5.142857143 is the mean.12.43956044 is the variance.3.526976104 is the standard deviation.


What is the standard deviation of 3 5 6 4 7 7 8?

2√(17/21), or about 1.8.


What is the Calculated Standard Deviation In The Following Data 13 12 9 15 11 16 17 8 12 7 12?

It is 3.045


What is the standard deviation of 5 5 5 7 9 9 9?

7


What is the standard deviation of 1 2 3 4 5 6 7 8 9 10?

Variance = 2.87 (approx).


How to calculate variance of 3 5 6 4 7 7 8?

To calculate variance, (population standard deviation squared), usually written as σ2, first find the mean. In this case, ( 3 + 5 + 6 + 4 + 7 + 7 +8 ) / 7 = 40/7 ≈ 5.7143 Then find the difference between every number and the mean. 5.7143 - 3 = 2.7143 5.7143 - 5 = 0.7143 5.7143 - 6 = -0.2857 5.7143 - 4 = 1.7143 5.7143 - 7 = -1.2857 5.7143 - 7 = -1.2857 5.7143 - 8 = -2.2857 Then square all of the differences. 7.3674 0.5102 0.0816 2.9388 1.6530 1.6530 5.2244 Then add the products together. 19.4284 Then divide by the number of numbers in the set of data (7) for the variance. σ2 ≈ 2.78 Since you are using the entire population, you use the population standard deviation instead of the sample standard deviation. In statistics, you are more likely to see sample standard deviation, but since there are only seven numbers in the data set, we use the population standard deviation.


How do you create data set with larger standard deviation?

Standard deviation is the square root of the sum of the squares of the deviations of each item from the mean, i.e. the square root of the variance. In order to increase the standard deviation, therefore, you need to increase the average deviation from the mean. There are many ways to do this. One is to move each item further away from the mean. For example, take the set [2, 4, 4, 4, 5, 5, 7, 9]. It has a mean of 5 and a standard deviation of 2.14. Multiply each item by 2.2 and subtract 5, giving the set [-1.3, 2.9, 2.9, 2.9, 5, 5, 9.2, 13.4], effectively moving each item 10% further away from the mean. This still has a mean of 5, but the standard deviation is 4.49.


A score set has a mean of 53 and a standard deviation of 5.6 If each score was multiplied by 0.9 and then had 7 added state the new standard deviation?

Standard deviation is the spread of the data. If each score has 7 added, this would not affect the spread of the data - it would be just as evenly spaced or clumped up, but 7 greater. The only thing that would affect the spread is multiplying every data point by 0.9. This makes distances between the data points 0.9 times as big, and thus makes the standard deviation 0.9 times as big. The standard deviation was 5.6, and so now is 5.6x0.9 = 5.04