Answer :
A sample of seven compact discs at the music store stated the performance times as lasting the following numbers of minutes for Beethoven's Ninth Symphony. To find the sample median, we need to arrange the performance times in ascending order:the sample standard deviation for Beethoven's Ninth Symphony is approximately 2.87 minutes.
65.4, 66.2, 66.9, 68.4, 68.6, 71.0, 71.9
The middle value of the ordered data is the sample median. Since there are 7 data points, the middle value will be the 4th value, which is 68.4 minutes. Therefore, the sample median for Beethoven's Ninth Symphony is 68.4 minutes.
To find the sample mean, we need to sum up all the performance times and divide the sum by the total number of data points:
66.9 + 66.2 + 71.0 + 68.6 + 65.4 + 68.4 + 71.9 = 468.4
Next, we divide the sum by 7 (the number of data points) to find the sample mean:
468.4 / 7 = 66.91
Therefore, the sample mean for Beethoven's Ninth Symphony is approximately 66.91 minutes.
To find the sample standard deviation, we first need to find the deviation of each data point from the sample mean. We then square each deviation, sum them up, divide by the total number of data points minus 1 (6 in this case), and finally take the square root of the result:
Deviation of 65.4: 65.4 - 66.91 = -1.51
Deviation of 66.2: 66.2 - 66.91 = -0.71
Deviation of 66.9: 66.9 - 66.91 = -0.01
Deviation of 68.4: 68.4 - 66.91 = 1.49
Deviation of 68.6: 68.6 - 66.91 = 1.69
Deviation of 71.0: 71.0 - 66.91 = 4.09
Deviation of 71.9: 71.9 - 66.91 = 4.99
Next, we square each deviation:
(-1.51)^2 = 2.28
(-0.71)^2 = 0.51
(-0.01)^2 = 0.00
(1.49)^2 = 2.22
(1.69)^2 = 2.86
(4.09)^2 = 16.73
(4.99)^2 = 24.90
Sum of squared deviations: 2.28 + 0.51 + 0.00 + 2.22 + 2.86 + 16.73 + 24.90 = 49.50
Now, we divide the sum by 6 (the number of data points minus 1) and take the square root of the result:
49.50 / 6 = 8.25
√8.25 ≈ 2.87
Therefore, the sample standard deviation for Beethoven's Ninth Symphony is approximately 2.87 minutes.
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