in this problem, you will use desmos to compute a few statistics.\n\n• open a new browser window to the…

in this problem, you will use desmos to compute a few statistics.\n\n• open a new browser window to the page: https://www.desmos.com/calculator.\n• enter the command below, by copying and pasting the data between the brackets:\n\na = paste data here \n\n• to compute the mean and median, enter the commands below:\n\nmean(a)\nmedian(a)\n\n• to compute the midrange of the data set, you will need the minimum and maximum values, which are computed in desmos by entering:\n\nmin(a)\nmax(a)\n\nthe heights of 60 randomly selected women are recorded below.\n\n{ 54, 54.7, 55.3, 55.6, 55.9, 56.4, 56.7, 57.5, 57.6, 57.8, 59.4, 59.8, 59.9, 60, 60.3, 61, 61, 61, 61, 61.2, 61.8, 62, 62.3, 62.5, 62.6, 62.7, 63, 63.4, 63.5, 63.6, 63.7, 63.7, 63.8, 63.8, 64.3, 64.4, 64.5, 64.6, 64.8, 65, 65, 65, 65.1, 65.1, 65.3, 65.4, 65.4, 66.1, 66.6, 66.6, 66.6, 67.3, 67.4, 68.4, 68.5, 69.1, 70.7, 71, 71.4, 72.2 }.\n\ngive the mean of the data set.\n\ngive the median of the data set.\n\ngive the midrange of the data set.\n\ngiven the relationship between the mean and median, what distribution shape likely?\n\nthe distribution will be roughly symmetric.\nthe distribution will probably be skewed to the right.\nthe distribution will probably be skewed to the left.
Answer
Explanation:
Step1: Recall mean - formula
The mean $\bar{x}$ of a data - set $x_1,x_2,\cdots,x_n$ is $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here $n = 60$.
Step2: Recall median - formula
For a data - set with $n = 60$ (an even number of data points), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data points.
Step3: Recall mid - range formula
The mid - range is $\frac{\min+\max}{2}$. First, find $\min$ and $\max$ of the data set. The minimum value of the data set is $54$ and the maximum value is $72.2$. So the mid - range is $\frac{54 + 72.2}{2}=\frac{126.2}{2}=63.1$.
Step4: Analyze distribution shape
If the mean is less than the median, the distribution is probably skewed to the left. If the mean is greater than the median, the distribution is probably skewed to the right. If the mean is approximately equal to the median, the distribution is roughly symmetric. Here, the mean is $62.385$ and the median is $63.75$. Since the mean is less than the median, the distribution will probably be skewed to the left.
Answer:
Part 1: $62.385$ Part 2: $63.75$ Part 3: $63.1$ Part 4: The distribution will probably be skewed to the left.