consider the data set 10,11,12,13,14,15,16,17,18. complete parts (a) through (c) below.\na. obtain the mean…

consider the data set 10,11,12,13,14,15,16,17,18. complete parts (a) through (c) below.\na. obtain the mean and median of the data.\nthe mean is 14. (type an integer or a decimal. do not round.)\nthe median is 14. (type an integer or a decimal. do not round.)\nb. replace the 18 in the data set by 108 and again compute the mean and median. decide which measure of center works better here, and explain your answer.\nthe mean is 24. (type an integer or a decimal. do not round.)\nthe median is 14. (type an integer or a decimal. do not round.)\nwhich center of measure works better here?\na. the median works better here since it is more typical of most of the data.\nb. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nc. the mean 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.\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 is not resistant to outliers. since the 18 in the data set was replaced by 108, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for the data.\nb. 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.\nc. 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.

consider the data set 10,11,12,13,14,15,16,17,18. complete parts (a) through (c) below.\na. obtain the mean and median of the data.\nthe mean is 14. (type an integer or a decimal. do not round.)\nthe median is 14. (type an integer or a decimal. do not round.)\nb. replace the 18 in the data set by 108 and again compute the mean and median. decide which measure of center works better here, and explain your answer.\nthe mean is 24. (type an integer or a decimal. do not round.)\nthe median is 14. (type an integer or a decimal. do not round.)\nwhich center of measure works better here?\na. the median works better here since it is more typical of most of the data.\nb. neither measure of center works for this data set. neither measure of center is typical of most of the data.\nc. the mean 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.\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 is not resistant to outliers. since the 18 in the data set was replaced by 108, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for the data.\nb. 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.\nc. 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.

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 data set $10,11,12,13,14,15,16,17,18$, $n = 9$ and $\sum_{i=1}^{9}x_{i}=10 + 11+12+13+14+15+16+17+18=\frac{(10 + 18)\times9}{2}=126$. So, $\bar{x}=\frac{126}{9}=14$.

Step2: Recall median formula

For a set of $n = 9$ (odd number of) data - points arranged in ascending order, the median is the $\left(\frac{n + 1}{2}\right)$-th value. Here, $\frac{9+1}{2}=5$ - th value, which is $14$.

Step3: Modify data set and recalculate mean

When we replace $18$ with $108$, the new sum $\sum_{i = 1}^{9}x_{i}=10+11+12+13+14+15+16+17+108=216$. The new mean $\bar{x}=\frac{216}{9}=24$.

Step4: Check median after data modification

Since the number of data - points is still $n = 9$ (odd) and the middle - valued data - point (the 5 - th value in ascending order) does not change, the median is still $14$.

Step5: Analyze measure of center

The median is more typical of most of the data as the outlier $108$ has a large effect on the mean, pulling it away from the central values of the original data set.

Step6: Analyze property of mean

The mean is not resistant to outliers. When an outlier (replacing $18$ with $108$) is introduced, the mean is pulled in the direction of the outlier and is no longer central or typical for the data.

Answer:

a. Mean: $14$, Median: $14$ b. Mean: $24$, Median: $14$, Answer: A. The median works better here since it is more typical of most of the data. c. Answer: A. The mean is not resistant to outliers. Since the 18 in the data set was replaced by 108, the mean is pulled in that direction due to that single observation, and is therefore neither central nor typical for the data.