Question

evaluating a histogram
the data set 80, 84, 79, 44, 87, 73, 89, 90, 82, 89, 93, 97, 77, and 71 is graphed below.
use the histogram to answer the questions.
what type of distribution is this?
skewed
what is the mean?
what is the median?
asure of center is the best choice for this data
81.071
82.5
83

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points. The data set is $60,84,79,44,87,73,89,90,82,89,93,97,77,71$. $n = 14$, and $\sum_{i=1}^{14}x_{i}=60 + 84+79+44+87+73+89+90+82+89+93+97+77+71=1135$. So, $\bar{x}=\frac{1135}{14}\approx81.071$.

Step2: Arrange data for median

Arrange the data in ascending order: $44,60,71,73,77,79,82,84,87,89,89,90,93,97$. Since $n = 14$ (an even number), the median is the average of the $\frac{n}{2}$ - th and $(\frac{n}{2}+1)$ - th ordered data points. $\frac{n}{2}=7$ and $\frac{n}{2}+1 = 8$. The 7 - th value is $82$ and the 8 - th value is $84$. So, the median $M=\frac{82 + 84}{2}=83$.

Answer:

Mean: $81.071$
Median: $83$