calculating mean, median, and outlier\njournalists try to use shorter or simpler words in their news stories…

calculating mean, median, and outlier\njournalists try to use shorter or simpler words in their news stories whenever possible. the box below shows the number of letters in each word taken from one sentence in a news article. what are the mean and median number of letters per word? what is the outlier?\nthe mean number of letters per word =\nthe median number of letters per word =\nthe outlier = \nthere is no outlier \nletters per word in a news article (12 - word sample)\n9 8\n2 2\n2 6\n11 1\n1 7\n8 3

calculating mean, median, and outlier\njournalists try to use shorter or simpler words in their news stories whenever possible. the box below shows the number of letters in each word taken from one sentence in a news article. what are the mean and median number of letters per word? what is the outlier?\nthe mean number of letters per word =\nthe median number of letters per word =\nthe outlier = \nthere is no outlier \nletters per word in a news article (12 - word sample)\n9 8\n2 2\n2 6\n11 1\n1 7\n8 3

Answer

Explanation:

Step1: List out the data

The data set from the stem - and - leaf plot is: 18, 11, 21, 22, 26, 37, 83, 98. First, we need to find the mean and median.

Step2: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 8$ and $\sum_{i=1}^{8}x_{i}=18 + 11+21+22+26+37+83+98=316$. So, $\bar{x}=\frac{316}{8}=39.5$.

Step3: Arrange data in ascending order

The ordered data set is: 11, 18, 21, 22, 26, 37, 83, 98.

Step4: Calculate the median

Since $n = 8$ (an even - numbered data set), the median $M=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}$. Here, $x_{4}=22$ and $x_{5}=26$, so $M=\frac{22 + 26}{2}=24$.

Step5: Check for outliers

We use the inter - quartile range (IQR) method. First, find the first quartile $Q_{1}$ and the third quartile $Q_{3}$. The lower half of the data is 11, 18, 21, 22 and $Q_{1}=\frac{18 + 21}{2}=19.5$. The upper half of the data is 26, 37, 83, 98 and $Q_{3}=\frac{37+83}{2}=60$. Then, $IQR = Q_{3}-Q_{1}=60 - 19.5 = 40.5$. The lower fence is $Q_{1}-1.5\times IQR=19.5-1.5\times40.5=19.5 - 60.75=-41.25$. The upper fence is $Q_{3}+1.5\times IQR=60 + 1.5\times40.5=60+60.75 = 120.75$. Since all the data points 11, 18, 21, 22, 26, 37, 83, 98 are within the fences, there is no outlier.

Answer:

The mean number of letters per word = 39.5 The median number of letters per word = 24 The outlier = There is no outlier