Question

（the text in the image is partially unclear, and the following is the recognizable part: there are some questions related to data analysis, including calculating the median, quartiles, etc. of two data sets, and comparing the number of large - value data in the two data sets.）

Explanation:

Step1: Calculate wild cats' mean

First, list the wild cats' speeds: 50, 40, 30, 45, 50, 25, 45, 55.
Mean $\bar{x}_1 = \frac{50+40+30+45+50+25+45+55}{8} = \frac{340}{8} = 42.5$

Step2: Calculate wild cats' variance

Compute squared differences from mean:
$(50-42.5)^2=56.25$, $(40-42.5)^2=6.25$, $(30-42.5)^2=156.25$, $(45-42.5)^2=6.25$,
$(50-42.5)^2=56.25$, $(25-42.5)^2=306.25$, $(45-42.5)^2=6.25$, $(55-42.5)^2=156.25$
Variance $s_1^2 = \frac{56.25+6.25+156.25+6.25+56.25+306.25+6.25+156.25}{8-1} = \frac{740}{7} \approx 105.7143$

Step3: Calculate wild cats' standard deviation

Standard deviation $\sigma_1 = \sqrt{\frac{740}{7}} \approx \sqrt{105.7143} \approx 10.28$

Step4: Calculate birds' mean

List the birds' speeds: 252, 195, 43, 53, 62, 29, 35, 28, 62, 25, 29, 30.
Mean $\bar{x}_2 = \frac{252+195+43+53+62+29+35+28+62+25+29+30}{12} = \frac{743}{12} \approx 61.9167$

Step5: Calculate birds' variance

Compute squared differences from mean:
$(252-61.9167)^2\approx36131.84$, $(195-61.9167)^2\approx17711.14$, $(43-61.9167)^2\approx357.84$,
$(53-61.9167)^2\approx79.51$, $(62-61.9167)^2\approx0.01$, $(29-61.9167)^2\approx1083.51$,
$(35-61.9167)^2\approx1363.14$, $(28-61.9167)^2\approx1148.38$, $(62-61.9167)^2\approx0.01$,
$(25-61.9167)^2\approx1362.88$, $(29-61.9167)^2\approx1083.51$, $(30-61.9167)^2\approx1018.75$
Sum of squared differences $\approx36131.84+17711.14+357.84+79.51+0.01+1083.51+1363.14+1148.38+0.01+1362.88+1083.51+1018.75=61340.58$
Variance $s_2^2 = \frac{61340.58}{12-1} \approx 5576.42$

Step6: Calculate birds' standard deviation

Standard deviation $\sigma_2 = \sqrt{5576.42} \approx 74.68$

Step7: Compare deviations

Difference in standard deviations: $10.28 < 74.68$, so wild cats' speeds deviate less.

Answer:

The standard deviation of the large wild cats is $\sigma \approx 10.28$
The standard deviation of the birds in flight is $\sigma \approx 74.68$
The standard deviations of the two data sets show that the fastest recorded speeds of large wild cats deviate less from the means than the fastest recorded speeds of birds in flight.