Find the standard deviation for each data set. Use the standard deviations to compare the pair of data sets. fastest recorded speeds of various large wild cats (miles per hour): 65 50 40 40 35 30 35 40 20 fasted recorded speeds of various birds in flight (miles per hour): 220 109 98 55 65 39 45 29 50 25 20 20 Question content area bottom Part 1 The standard deviation of the large wild cats is sigmaequalsenter your response here. (Round the final answer to the nearest hundredth as needed. Round all intermediate values to the nearest hundredth as needed.)
The standard deviation of the large wild cats is 11.15.
To calculate the standard deviation, follow these steps:
1. Find the mean of the data set for large wild cats:
(65 + 50 + 40 + 40 + 35 + 30 + 35 + 40 + 20) / 9 = 37.78
2. Subtract the mean from each data point, then square the result:
(65-37.78)^2 = 729.83
(50-37.78)^2 = 150.83
(40-37.78)^2 = 4.92
(40-37.78)^2 = 4.92
(35-37.78)^2 = 7.56
(30-37.78)^2 = 59.35
(35-37.78)^2 = 7.56
(40-37.78)^2 = 4.92
(20-37.78)^2 = 312.18
3. Find the sum of the squared differences:
729.83 + 150.83 + 4.92 + 4.92 + 7.56 + 59.35 + 7.56 + 4.92 + 312.18 = 1282.77
4. Divide the sum by the number of data points minus 1, then take the square root:
√(1282.77 / 8) ≈ 11.15
Therefore, the standard deviation for the large wild cats data set is 11.15.