ANSWER CHOICE C
We have been given two box plots representing attendance at a local movie theater and high school basketball games. WE are asked to choose the correct option about the measure the spread of the data represented by box plots.
Since we know that Standard deviation is not a good measure of spread in highly skewed data. IQR is better measure of spread for highly skewed data.
We can see that box plot representing attendance at movies is right skewed, therefore, mean will be greater than median and SD will not be a good measure of spread.
Box plot representing basket-ball games attendance is left skewed left,therefore, mean will be less than median and SD will not be a good measure of spread.
Our both box plots are highly skewed, therefore, the IQR is the best measurement of spread for games and movies and option C is the correct choice.
Answer:
1/2 or 0.5 is the correct answer of this question
M1 = m4 = 50 (A) is the answer
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Thanks for the points.
