Answer:
Step-by-step explanation:
The given data set for type 1 of birds is:
Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}
Mean of the type 1 data is:
\(Mean=\frac{18+24+20+22+26}{5}=\frac{110}{5}=22\)
Data \((x-{\overline{x})^2\)
18 16
24 4
20 4
22 0
26 16
Now, mean average of squares is:
\(m=\frac{40}{5}=8\)
Standard deviation=\(SD=\sqrt{8}=2.828\)
Now, the difference of mean and its standard deviation of type 1 data set is:
=22-2.828
Difference =19.172
The given data set for type 2 of birds is:
Type 2: {24, 21, 19, 26, 30}
Mean of the type 2 data is:
\(Mean=\frac{24+21+19+26+30}{5}=\frac{120}{5}=24\)
Data \((x-{\overline{x})^2\)
24 0
21 9
19 25
26 4
30 36
Now, mean average of squares is:
\(m=\frac{74}{5}\)
\(m=14.8\)
Standard deviation=\(SD=\sqrt{14.8}=3.84\)
Now, the difference of mean and its standard deviation of type 2 data set is:
=24-3.84
Difference=20.16
Since, the difference of mean and standard deviation of both type 1 and type 2 data set is different, therefore, Type 1 birds and Type 2 birds do not have similar wingspan distributions.
Hence, option C is correct.
