Respuesta :
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
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: 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.
Step-by-step explanation:
Given data : {1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 8}
[tex]\text{Mean }=\dfrac{\text{Sum of all data values}}{\text{Number of data values}}\\\\\Rightarrow\ \text{Mean }=\dfrac{56}{16}=3.5[/tex]
Now, data is already arranged in ascending order.
Since number of observations (16) is even.
Median = Mean of [tex](\dfrac{16}{2}\text{ and } \dfrac{16}{2}+1)^{th}\text{term}[/tex]
=Mean of [tex](8^{th}\text{ and }9^{th} \text{term})[/tex]
⇒Median[tex]=\dfrac{3+3}{2}=3[/tex]
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, [tex]\text{Mean }=\dfrac{68}{18}=3.77777\approx3.8[/tex]
Again number of observations (18) is even.
Median = Mean of [tex](\dfrac{18}{2}\text{ and } \dfrac{18}{2}+1)^{th}\text{term}[/tex]
=Mean of [tex](9^{th}\text{ and }10^{th} \text{term})[/tex]
⇒Median[tex]=\dfrac{3+4}{2}=3.5[/tex]
Difference of the mean and the median for the new data set is [tex]3.78-3.5=0.28[/tex]