Question

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 40 40 50 45 30 25 45 15\
fastest recorded speeds of various birds in flight (miles per hour)\
217 109 93 55 62 39 55 28 52 25 20 30\
\
the standard deviation of the large wild cats is \\(sigma = \square\\)\
(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 birds in flight is \\(sigma = \square\\)\
(round the final answer to the nearest hundredth as needed. round all intermediate values to the nearest hundredth as needed.)\
the standard deviations of the two data sets show that the fastest recorded speeds of large wild cats deviate \\(\square\\) from the mean than the fastest recorded speeds of birds in flight

Explanation:

Step1: Calculate wild cats' mean

First, find the mean of the wild cats' speeds.
$\mu_1 = \frac{60 + 40 + 50 + 45 + 30 + 25 + 45 + 15}{8} = \frac{310}{8} = 38.75$

Step2: Compute wild cats' squared deviations

Find squared difference from mean for each value.
$(60-38.75)^2=451.56$, $(40-38.75)^2=1.56$, $(50-38.75)^2=126.56$, $(45-38.75)^2=39.06$,
$(30-38.75)^2=76.56$, $(25-38.75)^2=189.06$, $(45-38.75)^2=39.06$, $(15-38.75)^2=564.06$

Step3: Calculate wild cats' variance

Average the squared deviations (population variance).
$\sigma_1^2 = \frac{451.56+1.56+126.56+39.06+76.56+189.06+39.06+564.06}{8} = \frac{1487.48}{8} = 185.935$

Step4: Find wild cats' standard deviation

Take square root of variance.
$\sigma_1 = \sqrt{185.935} \approx 13.64$

Step5: Calculate birds' mean

Find the mean of the birds' speeds.
$\mu_2 = \frac{217 + 109 + 93 + 55 + 62 + 39 + 55 + 28 + 52 + 25 + 20 + 30}{12} = \frac{785}{12} \approx 65.42$

Step6: Compute birds' squared deviations

Find squared difference from mean for each value.
$(217-65.42)^2=22976.49$, $(109-65.42)^2=1899.22$, $(93-65.42)^2=760.66$, $(55-65.42)^2=108.58$,
$(62-65.42)^2=11.69$, $(39-65.42)^2=698.02$, $(55-65.42)^2=108.58$, $(28-65.42)^2=1400.26$,
$(52-65.42)^2=180.10$, $(25-65.42)^2=925.38$, $(20-65.42)^2=1258.58$, $(30-65.42)^2=1254.58$

Step7: Calculate birds' variance

Average the squared deviations (population variance).
$\sigma_2^2 = \frac{22976.49+1899.22+760.66+108.58+11.69+698.02+108.58+1400.26+180.10+925.38+1258.58+1254.58}{12} = \frac{31882.14}{12} \approx 2656.85$

Step8: Find birds' standard deviation

Take square root of variance.
$\sigma_2 = \sqrt{2656.85} \approx 51.54$

Step9: Compare standard deviations

Compare the two standard deviation values.
$13.64 < 51.54$, so wild cats' speeds deviate less.

Answer:

The standard deviation of the large wild cats is $\sigma = 13.64$
The standard deviation of the birds in flight is $\sigma = 51.54$
The standard deviations of the two data sets show that the fastest recorded speeds of large wild cats deviate less from the mean than the fastest recorded speeds of birds in flight