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):
60  55  30  40  40  25  25  45  15
fasted recorded speeds of various birds in flight​ (miles per​ hour):
218  103  93  58  67  39  50  28  55  30  30  25

Bot GPT 3.5 Bot GPT 3.5 answered
1 month ago

To find the standard deviation for each data set, we first need to calculate the mean of each data set.

For the data set of fastest recorded speeds of various large wild cats:
Mean = (60 + 55 + 30 + 40 + 40 + 25 + 25 + 45 + 15) / 9
Mean = 295 / 9
Mean = 32.78 (rounded to 2 decimal places)

Next, we calculate the squared difference between each data point and the mean, then find the average of those squared differences to get the variance, and finally, take the square root of the variance to get the standard deviation.

Variance = [(60-32.78)^2 + (55-32.78)^2 + (30-32.78)^2 + (40-32.78)^2 + (40-32.78)^2 + (25-32.78)^2 + (25-32.78)^2 + (45-32.78)^2 + (15-32.78)^2] / 9
Variance = [943.52 + 506.74 + 9.23 + 61.04 + 61.04 + 63.53 + 63.53 + 156.05 + 304.37] / 9
Variance = 2088.05 / 9
Variance = 232

Standard Deviation = √232
Standard Deviation = 15.23 (rounded to 2 decimal places)

Now, repeat the same process for the data set of fastest recorded speeds of various birds in flight:

Mean = (218 + 103 + 93 + 58 + 67 + 39 + 50 + 28 + 55 + 30 + 30 + 25) / 12
Mean = 726 / 12
Mean = 60.50 (rounded to 2 decimal places)

Variance = [(218-60.50)^2 + (103-60.50)^2 + (93-60.50)^2 + (58-60.50)^2 + (67-60.50)^2 + (39-60.50)^2 +(50-60.50)^2 + (28-60.50)^2 + (55-60.50)^2 + (30-60.50)^2 + (30-60.50)^2 + (25-60.50)^2] / 9
Variance = [8117.50 + 1365.25 + 815.62 + 5.62 + 43.87 + 430.59 + 117.62 + 1195.52 + 31.50 + 900.25 + 900.25 + 1382.12] / 12
Variance = 15672.51 / 12
Variance = 1306

Standard Deviation = √1306
Standard Deviation = 36.14 (rounded to 2 decimal places)

Comparing the standard deviations of the two data sets:
The large wild cat data set has a standard deviation of 15.23, while the bird in flight data set has a standard deviation of 36.14. This indicates that the data in the bird in flight data set is more spread out or varied compared to the data in the large wild cat data set.

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