Here are the prices for biweekly lawn mowing services for 12 local companies:
15, 15, 20, 30, 30, 35, 40, 40, 40, 40, 50, 50, 50
Which measure of center—mean or median—best represents the shape of the distribution?
The median would be the best measure of center for this distribution because there are some extreme values (such as $15 and $50) that could skew the mean. The median would give a better representation of the typical price for biweekly lawn mowing services in this area.
To determine whether the mean or median best represents the shape of the distribution, we need to consider the skewness of the data.
The given prices for biweekly lawn mowing services are: 15, 15, 20, 30, 30, 35, 40, 40, 40, 40, 50, 50, 50.
First, let's calculate the mean and median:
Mean: To calculate the mean, we sum up all the prices and divide by the total number of prices. The sum of the prices is 475. Dividing this by 13 (the total number of prices) gives us a mean of approximately 36.54.
Median: To calculate the median, we arrange the data in ascending order and find the middle value. In this case, there are 13 values, so the middle value is the 7th price, which is 40.
Now, let's consider the shape of the distribution. Since the data is not given in a frequency distribution, we'll assess the shape visually:
15, 15, 20, 30, 30, 35, 40, 40, 40, 40, 50, 50, 50
The data appears to be right-skewed or positively skewed, as there are more prices clustered towards the lower end of the range with a few higher values.
Given the shape of the data, the median is a better measure of center to represent the distribution. It is less affected by extreme values in the upper end (such as the values of 50) and gives a better representation of the typical value in the dataset. The median of 40 falls within the cluster of prices that are closer together and represents a more central value compared to the mean of 36.54.
Therefore, in this case, the median best represents the shape of the distribution.