Question

Skew of Distribution

A comic-strip writer churns out different numbers of comic strips each day. The writer logged the number of comic strips he wrote each day for 16 days and sorted the data in ascending order to create this data set.

{1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 8}

The skew in the distribution is . If the writer writes for two more days and creates 5 and 7 comic strips, respectively, the difference of the mean and the median for the new data set is

Answer 1

For the answer to the question above asking what the data set is, if the writer writes for two more days and creates 5 and 7 comic strips, respectively, the difference of the mean and the median?
For the 1st data set: 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 8

mean: 3.5
median: 3

2nd Data set: 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 8

mean: 3.78
median: 3.5

difference in mean: 3.78 - 3.5 = 0.28
difference in median: 3.5 - 3 = 0.50

Answer 2

Answer: The skew in the distribution is positive. If the writer writes for two more days and creates 5 and 7 comic strips, respectively, the difference of the mean and the median for the new data set is 0.28.

Explanation:

Given data : {1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 8}

`\text{Mean }=\frac{\text{Sum of all data values}}{\text{Number of data values}}\\\\\Rightarrow\ \text{Mean }=(56)/(16)=3.5`

Now, data is already arranged in ascending order.

Since number of observations (16) is even.

Median = Mean of
`((16)/(2)\text{ and } (16)/(2)+1)^(th)\text{term}`

=Mean of
`(8^(th)\text{ and }9^(th) \text{term})`

⇒Median
`=(3+3)/(2)=3`

Since, Median < Mean

Then , the data is positively skewed.

The new data set : {1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 8,5,7}

Number of data values :18

Now,
`\text{Mean }=(68)/(18)=3.77777\approx3.8`

Again number of observations (18) is even.

Median = Mean of
`((18)/(2)\text{ and } (18)/(2)+1)^(th)\text{term}`

=Mean of
`(9^(th)\text{ and }10^(th) \text{term})`

⇒Median
`=(3+4)/(2)=3.5`

Difference of the mean and the median for the new data set is
`3.78-3.5=0.28`