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Question 1 of 2
The height of a group of 5 dogs and a group of 5 cats are collected in centimetres. Compare the mean for dogs ¯xD and the mean for cats ¯xC.
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¯xD= dogs mean
¯xC= cats mean
Use the mean formula to find the average height of the dogs.
¯xD |
= |
sum of datanumber of data |
|
|
= |
20+30+40+50+605 |
Substitute values |
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|
= |
2005 |
|
|
= |
40 cm |
Simplify |
Use the mean formula to find the average height of the cats.
¯xC |
= |
sum of datanumber of data |
|
|
= |
38+39+40+41+425 |
Substitute values |
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|
= |
2005 |
|
|
= |
40 cm |
Simplify |
The mean of both groups is 40 cm even though their heights are not the same.
This only means that both groups have their data centralized or averaged at 40 cm.
¯xD=40 cm
¯xC=40 cm
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Question 2 of 2
The height of a group of 5 dogs and a group of 5 cats are collected in centimetres. Find and compare the standard deviation of both groups.
The mean of both groups is given: ¯¯¯X=40
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To solve for the standard deviation for the height of the dogs, label the given values needed for the formula.
X values (heights)=20,30,40,50,60
Mean ¯¯¯X=40
Number of dogs: n=5
Subsitute into the standard deviation formula
S |
= |
√∑(X−¯X)2n |
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|
= |
√(20−40)2+(30−40)2+(40−40)2+(50−40)2+(60−40)25 |
Substitute values |
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= |
√(−20)2+(−10)2+(0)2+(10)2+(20)25 |
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= |
√400+100+0+100+4005 |
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= |
√10005 |
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= |
√200 |
Sd |
= |
14.1 |
To solve for the standard deviation for the height of the cats, label the given values needed for the formula.
X values (heights)=38,39,40,41,42
Mean ¯¯¯X=40
Number of cats: n=5
Subsitute into the standard deviation formula
S |
= |
√∑(X−¯X)2n |
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|
= |
√(38−40)2+(39−40)2+(40−40)2+(41−40)2+(42−40)2)5 |
Substitute values |
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= |
√(−2)2+(−1)2+(0)2+(1)2+(2)25 |
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= |
√4+1+0+1+45 |
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= |
√105 |
Simplify |
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|
= |
√2 |
Sc |
= |
1.4 |
Sd is greater than Sc.
This means that the height of the dogs is more dispersed than the height of the cats.
Sd=14.1cm
Sc=1.4cm