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Angela is studying two types of birds. She measures the wingspan, in inches, of five Type 1 birds and five Type 2 birds.

What can she infer about the wingspans of the two types of birds?

Type 1: {18, 24, 20, 22, 26}
Type 2: {24, 21, 19, 26, 30}

A.
Type 1 and Type 2 birds have similar wingspan distributions.

B.
Type 1 and Type 2 birds have somewhat similar wingspan distributions.

C.
Type 1 birds and Type 2 birds do not have similar wingspan distributions.

D.
Type 1 birds and Type 2 birds have identical wingspan distributions.

Sagot :

boffeemadrid

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:

[tex]Mean=\frac{18+24+20+22+26}{5}=\frac{110}{5}=22[/tex]

Data                                                         [tex](x-{\overline{x})^2[/tex]

18                                                                    16

24                                                                   4

20                                                                   4

22                                                                   0

26                                                                   16

Now, mean average of squares is:

[tex]m=\frac{40}{5}=8[/tex]

Standard deviation=[tex]SD=\sqrt{8}=2.828[/tex]

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:

[tex]Mean=\frac{24+21+19+26+30}{5}=\frac{120}{5}=24[/tex]

Data                                                        [tex](x-{\overline{x})^2[/tex]

24                                                                     0

21                                                                      9

19                                                                     25

26                                                                     4

30                                                                    36

Now, mean average of squares is:

[tex]m=\frac{74}{5}[/tex]

[tex]m=14.8[/tex]

Standard deviation=[tex]SD=\sqrt{14.8}=3.84[/tex]

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.

craftypandacute

Answer:

Type 1 and Type 2 birds have similar wingspan distributions.

Step-by-step explanation:

View image craftypandacute
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