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. both centers of measure work equally well here. they are both 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. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nc. for the data set in part (b), the mean is neither central nor typical for the data. the lack of what property of the mean accounts for this result?\na. the mean has a lower probability of being observed. whereas observations in the data set have been observed and are likely to be obtained in repeated sampling, the mean is not necessarily observable, and is therefore neither central nor typical for the data.\nb. the mean is not constant. since the mean changes depending on the observations in the data set and/or the number of observations, the mean cannot be representative of the underlying population.\nc. the mean is not resistant to outliers. since the 19 in the data set was replaced by 109, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for 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. both centers of measure work equally well here. they are both 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. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nc. for the data set in part (b), the mean is neither central nor typical for the data. the lack of what property of the mean accounts for this result?\na. the mean has a lower probability of being observed. whereas observations in the data set have been observed and are likely to be obtained in repeated sampling, the mean is not necessarily observable, and is therefore neither central nor typical for the data.\nb. the mean is not constant. since the mean changes depending on the observations in the data set and/or the number of observations, the mean cannot be representative of the underlying population.\nc. the mean is not resistant to outliers. since the 19 in the data set was replaced by 109, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for 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. For the original 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 $\bar{x}=\frac{135}{9}=15$. 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 itself, and the $\left(\frac{9+1}{2}\right)$-th element is the 5 - th element, which is 15.

Step2: Calculate mean and median after replacement

After replacing 19 with 109, $n = 9$ still, and $\sum_{i = 1}^{9}x_{i}=11+12+13+14+15+16+17+18+109=135 - 19+109=225$. So the new mean $\bar{x}=\frac{225}{9}=25$. The ordered set is $11,12,13,14,15,16,17,18,109$, and the median (the 5 - th element) is 15. The median is better here because the outlier 109 has a large effect on the mean, pulling it away from the typical values of the data, while the median remains the same and is more representative of the majority of the data values.

Step3: Identify the property of the mean

The mean is not resistant to outliers. When an outlier (109 instead of 19) is introduced, the mean is pulled in the direction of the outlier, making it non - central and non - typical for the data.

Answer:

a. Mean: 15, Median: 15 b. Mean: 25, Median: 15, Answer: C. The median works better here since it is more typical of most of the data. c. Answer: C. The mean is not resistant to outliers. Since the 19 in the data set was replaced by 109, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for the data.