Question

multiple choice: select the best answer for exercises 123–126.

1. if a distribution is skewed to the right with no outliers,

which expression is correct?
(a) mean < median \t\t(d) mean > median
(b) mean ≈ median \t\t(e) we cant tell without
(c) mean = median \t\t\texamining the data.

1. the scores on a statistics test had a mean of 81 and

a standard deviation of 9. one student was absent
on the test day, and his score wasn’t included in the
calculation. if his score of 84 was added to the distri-
bution of scores, what would happen to the mean and
standard deviation?
(a) mean will increase, and standard deviation will
\tincrease.
(b) mean will increase, and standard deviation will
\tdecrease.
(c) mean will increase, and standard deviation will stay
\tthe same.
(d) mean will decrease, and standard deviation will
\tincrease.
(e) mean will decrease, and standard deviation will
\tdecrease.

Explanation:

Question 123

Step1: Recall skewed right distribution

In a right - skewed (positively skewed) distribution with no outliers, the tail is on the right side. The mean is pulled in the direction of the tail (to the right) because it is sensitive to extreme values (even though there are no outliers, the longer tail on the right affects it). The median is the middle value and is less affected by the skewness. So, the mean is greater than the median.

Step1: Analyze the effect on the mean

The original mean of the test scores is 81. The new score added is 84, which is greater than the original mean (81). The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). When we add a score \(x_{n + 1}=84\) (which is greater than the original mean), the new sum \(\sum_{i=1}^{n + 1}x_{i}=\sum_{i = 1}^{n}x_{i}+84\) and the new number of data points \(n+1\). Since 84>81, the new mean \(\bar{x}_{new}=\frac{\sum_{i = 1}^{n}x_{i}+84}{n + 1}\) will be greater than the original mean of 81.

Step2: Analyze the effect on the standard deviation

The formula for the standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\) (for sample standard deviation) or \(\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}}\) (for population standard deviation). The new score is 84, and the original mean is 81. The difference between the new score and the original mean is \(84 - 81=3\). When we add this score, we first calculate the new mean (which is slightly greater than 81, say \(\bar{x}_{new}\)). The deviation of the new score from the new mean \((84-\bar{x}_{new})\) will be less than 3 (because \(\bar{x}_{new}>81\)). The sum of squared deviations \(\sum(x_{i}-\bar{x})^{2}\) will increase by \((84 - \bar{x}_{new})^{2}\), but the number of data points also increases. However, since the new score is closer to the new mean than it is to the original mean (because the new mean is pulled up towards 84 from 81), the increase in the sum of squared deviations is relatively small compared to the increase in the number of data points. In general, when we add a data point that is closer to the mean (after recalculating the mean) than the spread of the original data, the standard deviation decreases. Since 84 is closer to the original mean (81) than being an extreme value, adding it will decrease the standard deviation.

Answer:

(d) mean > median

Question 124