Answer:
The data for April is negatively skewed, hence, most possibly the mean of the distribution is less than the median which is, 67.
The data for October is positively skewed, hence, most possibly the mean of the distribution is greater than the median which is, 60.
But, there is possibility for otherwise also in both of the cases, so there is actually no way of predicting what the actual means are.
Step-by-step explanation:
The Bowley's measure of skewness for the data on the box-plot showing average daily temperatures in April for a U. S. city is given by,
[tex]\frac {Q_{3} + Q_{1} - 2Q_{2}}{Q_{3} - Q_{1}}[/tex]
= [tex]\frac {71 + 60 - 2 \times 67}{71 - 60}[/tex]
= [tex]\frac {- 3}{11}[/tex] < 0
Hence, the data is negatively skewed, hence, most possibly the mean of the distribution is less than the median which is, 67.
The Bowley's measure of skewness for the data on the box-plot showing average daily temperatures in October for a U. S. city is given by,
[tex]\frac {Q_{3} + Q_{1} - 2Q_{2}}{Q_{3} - Q_{1}}[/tex]
= [tex]\frac {75 + 50 - 2 \times 60}{75 - 50}[/tex]
= [tex] 0.2[/tex] > 0
Hence, the data is positively skewed, hence, most possibly the mean of the distribution is greater than the median which is, 60.
Answer:
There is no way of telling what the means are.
Step-by-step explanation:
I took the test and got it right
