Calculate the mean, x⎯⎯⎯, and standard deviation, , for the data set.
Sample Value
1 6.016
2 6.029
3 6.026
4 6.025
5 6.018
6 6.015
The Correct Answer and Explanation is:
The mean ((\bar{x})) of the given data set is 6.0215, and the standard deviation ((s)) is approximately 0.00589.
Explanation:
- Mean Calculation:
The mean, also known as the arithmetic average, is calculated as the sum of all values divided by the number of values in the data set:
[
\bar{x} = \frac{\sum x_i}{n}
]
Here:
[
x_1 = 6.016, \, x_2 = 6.029, \, x_3 = 6.026, \, x_4 = 6.025, \, x_5 = 6.018, \, x_6 = 6.015
]
Summing these values gives:
[
\sum x_i = 6.016 + 6.029 + 6.026 + 6.025 + 6.018 + 6.015 = 36.129
]
Dividing by the number of samples ((n = 6)):
[
\bar{x} = \frac{36.129}{6} = 6.0215
] - Standard Deviation Calculation:
The standard deviation measures the dispersion of the data from the mean. For a sample, the formula is:
[
s = \sqrt{\frac{\sum (x_i – \bar{x})^2}{n-1}}
]
Subtracting the mean from each data point gives deviations:
[
\text{Deviations: } -0.0055, 0.0075, 0.0045, 0.0035, -0.0035, -0.0065
]
Squaring these deviations and summing them gives:
[
\sum (x_i – \bar{x})^2 = 0.00003025 + 0.00005625 + 0.00002025 + 0.00001225 + 0.00001225 + 0.00004225 = 0.0001735
]
Dividing by (n-1 = 5) and taking the square root:
[
s = \sqrt{\frac{0.0001735}{5}} = \sqrt{0.0000347} \approx 0.00589
]
Significance:
- The mean represents the central tendency of the data, providing an average value.
- The standard deviation shows the variability; a smaller value like 0.00589 indicates that the data points are closely clustered around the mean. This is important in assessing consistency in measurements.