the accompanying data represent the weights (in grams) of a simple random sample of 10 m&m plain candies…

the accompanying data represent the weights (in grams) of a simple random sample of 10 m&m plain candies. determine the shape of the distribution of weights of m&ms by drawing a frequency histogram. find the mean and median. which measure of central tendency better describes the weight of a plain m&m? click the icon to view the candy weight data. the mean is x = 0.871 gram(s). (type an integer or a decimal. do not round.) the median is m = 0.84 gram(s). (type an integer or a decimal. do not round)

Answer

Explanation:

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are data - points and $n$ is the number of data - points. We assume the data set is $x_1,x_2,\cdots,x_{10}$. But since the mean $\bar{x}=0.871$ is already given, we don't need to calculate it from scratch.

Step2: Recall median formula

For a data set of size $n = 10$ (an even - numbered data set), the median $M$ is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data points. Since the median $M = 0.84$ is already given, we don't need to calculate it from scratch.

Step3: Analyze histogram for shape

Looking at the histograms:

A symmetric histogram has a similar shape on both sides of the center.
A skewed - right histogram has a long tail on the right side.
A skewed - left histogram has a long tail on the left side. The histogram in option C appears to be symmetric as the bars on either side of the center have a similar distribution. And when the distribution is symmetric, the mean and median are close in value. Here, the mean is $0.871$ and the median is $0.84$, which is consistent with a symmetric distribution.