describing and comparing distributions\nname name redacted\n1 some students purchased pumpkins for a carving…
describing and comparing distributions\nname name redacted\n1 some students purchased pumpkins for a carving contest. before the contest began, they weighed the pumpkins. the weights in pounds are shown here, along with a histogram of the data.\n3.6 4.0 9.6 14.0 11.0 12.4 13.0 2.0 6.0 6.6 15.0 5.4\n12.7 6.0 2.8 9.6 4.0 6.1 5.4 11.9 5.4 31.0 33.0\na. calculate the mean weight of the pumpkins.\n9.93\nb. find the median weight of the pumpkins.\n6.6\nc. would you use the mean or the median to summarize the typical weight of a pumpkin in this contest? explain\nmedian\nd. describe the distribution.
Answer
Explanation:
Step1: List the data values
3.6, 4.0, 9.6, 14.0, 11.0, 12.4, 13.0, 2.0, 6.0, 6.6, 15.0, 5.4, 12.7, 6.0, 2.5, 9.6, 4.0, 6.1, 5.4, 11.9, 5.4, 31.0, 33.0
Step2: Calculate the sum of data values
$3.6 + 4.0+9.6 + 14.0+11.0+12.4+13.0+2.0+6.0+6.6+15.0+5.4+12.7+6.0+2.5+9.6+4.0+6.1+5.4+11.9+5.4+31.0+33.0 = 228.4$
Step3: Count the number of data values
There are 23 data - values.
Step4: Calculate the mean
Mean=$\frac{228.4}{23}\approx9.93$
Step5: Arrange the data in ascending order
2.0, 2.5, 3.6, 4.0, 4.0, 5.4, 5.4, 5.4, 6.0, 6.0, 6.1, 6.6, 9.6, 9.6, 11.0, 11.9, 12.4, 12.7, 13.0, 14.0, 15.0, 31.0, 33.0
Step6: Find the median
Since there are 23 (an odd number) data - values, the median is the 12th value. The 12th value is 6.6.
Step7: Decide between mean and median
The data has outliers (31.0 and 33.0). The mean is affected by outliers, while the median is not. So, the median is a better measure to summarize the typical weight.
Step8: Describe the distribution
The distribution is skewed right because there are a few large values (outliers) that pull the right - tail of the distribution.
Answer:
a. 9.93 b. 6.6 c. median, because of outliers d. skewed right