consider the data set 11,12,13,14,15,16,17,18,19. complete parts (a) through (c) below.\na. obtain the mean…

consider the data set 11,12,13,14,15,16,17,18,19. complete parts (a) through (c) below.\na. obtain the mean and median of the data.\nthe mean is 15.\n(type an integer or a decimal. do not round.)\nthe median is 15.\n(type an integer or a decimal. do not round.)\nb. replace the 19 in the data set by 109 and again compute the mean and median. decide which measure of center works better here, and explain your answer.\nthe mean is 25.\n(type an integer or a decimal. do not round.)\nthe median is 15.\n(type an integer or a decimal. do not round.)\nwhich center of measure works better here?\na. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nb. the mean works better here since it is more typical of most of the data.\nc. the median works better here since it is more typical of most of the data.\nd. both centers of measure work equally well here. they are both typical of most of the data.

consider the data set 11,12,13,14,15,16,17,18,19. complete parts (a) through (c) below.\na. obtain the mean and median of the data.\nthe mean is 15.\n(type an integer or a decimal. do not round.)\nthe median is 15.\n(type an integer or a decimal. do not round.)\nb. replace the 19 in the data set by 109 and again compute the mean and median. decide which measure of center works better here, and explain your answer.\nthe mean is 25.\n(type an integer or a decimal. do not round.)\nthe median is 15.\n(type an integer or a decimal. do not round.)\nwhich center of measure works better here?\na. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nb. the mean works better here since it is more typical of most of the data.\nc. the median works better here since it is more typical of most of the data.\nd. both centers of measure work equally well here. they are both typical of most of the data.

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.

Step2: Calculate original mean

For the data set $11,12,13,14,15,16,17,18,19$, $n = 9$, and $\sum_{i=1}^{9}x_{i}=11 + 12+13+14+15+16+17+18+19=\frac{(11 + 19)\times9}{2}=135$. So the mean $\bar{x}=\frac{135}{9}=15$. The median of a set with $n = 9$ (odd number of elements) is the $\left(\frac{n + 1}{2}\right)$-th ordered element. The ordered set is the given set, and the $\left(\frac{9+1}{2}\right)$-th element is the 5 - th element, which is 15.

Step3: Replace data - point and calculate new mean

Replace 19 with 109. Now $\sum_{i = 1}^{9}x_{i}=11+12+13+14+15+16+17+18+109=135 - 19+109=225$. The new mean is $\frac{225}{9}=25$. The median remains the same (15) because the middle - value of the ordered set does not change when only the largest value is replaced.

Step4: Analyze measure of center

The median is more typical of most of the data since most of the data values are in the range of 11 - 18. The mean is affected by the outlier (109) and is not representative of the majority of the data.

Answer:

a. Mean: 15, Median: 15 b. Mean: 25, Median: 15 c. C. The median works better here since it is more typical of most of the data.