Ramiro did a survey of the number of pets owned by his classmates, with the following result

What was the mean Deviation of the number of pets?

A-1.02
B-1.053
C-1.087
D-1.12

Ramiro did a survey of the number of pets owned by his classmates with the following result What was the mean Deviation of the number of pets A102 B1053 C1087 D class=

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C is the answer .....

The mean deviation of the number of pets Ramiro surveyed as shown in the table attached below is: D. 1.12

To find the mean deviation of the number of pets, the following steps would be taken:

  • Find the mean, which is sum of each frequency and score (fx) divided by the total number of pets. [tex]\mathbf{(\frac{\sum fx}{n} )}[/tex]
  • Subtract the mean from each score, [tex]\mathbf{( x - \bar x)}[/tex]
  • Sum the deviations together and divide by total number of pets.
  • The formula for the mean deviation is given as: [tex]\mathbf{\frac{\sum f* |x - \bar x|}{n} }[/tex]

Mean deviation as solved in the table attached in the image below would be:

[tex]\mathbf{\frac{33.6}{30} = 1.12}[/tex]

Therefore, the mean deviation of the number of pets Ramiro surveyed as shown in the table attached below is: D. 1.12

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https://brainly.com/question/24185612

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