the box plot below displays the average number of apple pies consumed in eating contests across a particular…

Explanation:

Step1: Recall Mean and Median Properties

Mean is the average ((\frac{\sum x}{n})), sensitive to outliers. Median is the middle value, less sensitive to outliers (depends on middle data, not extreme values).

Step2: Analyze Outliers' Effect on Mean

Outliers (high and low) skew the mean. Removing both outliers (one high, one low) – the high outlier increases the mean, the low outlier decreases it. But in the box plot, the high outlier is much higher (professional contest) and low outlier much lower (children’s contest). Wait, no – wait, the box plot: the median line is inside the box. Removing outliers: the median is determined by the middle of the remaining data (the box’s middle line, which is the median of the original data excluding outliers? Wait, no – the box plot’s median is the median of all data (including outliers). Wait, no: box plot components: median (line in box), Q1, Q3, whiskers (to non - outlier data), outliers (dots). So the original median is the middle of all data (including outliers). But when we remove outliers, the new data set is the non - outlier data. The median of the non - outlier data: since the box (Q1, median, Q3) is based on the interquartile range (non - outlier middle 50%? No, box plot: median is the median of all data, Q1 and Q3 are quartiles of all data, whiskers extend to the furthest non - outlier data, outliers are beyond 1.5*IQR. So the original median is the middle value of all data (including outliers). When we remove outliers, the new data set is the data without the outliers. The median of the new data: since the outliers are at the extremes, removing them doesn't change the middle position of the non - outlier data (the median of the non - outlier data is the same as the original median, because the original median was in the middle of the non - outlier data? Wait, no. Wait, the box’s median line is the median of all data. But the non - outlier data: the median of the non - outlier data would be the same as the original median if the number of outliers is even (one high, one low) and the original median is in the middle of the non - outlier data. For the mean: the high outlier is a large value, the low outlier is a small value. Removing a large value (decreases the sum) and a small value (increases the sum). But the high outlier is much larger (professional contest: eats many pies) and the low outlier is much smaller (children’s contest: eats few). Wait, no – the high outlier is above the upper whisker, low below lower whisker. So the high outlier is a large number, low is a small number. Removing the high outlier (which was increasing the mean) and the low outlier (which was decreasing the mean). But which effect is stronger? Wait, no – wait, the original mean: sum includes high (H) and low (L) outliers. When we remove H and L, the new sum is (original sum - H - L), new n is (original n - 2). The median: original median is the middle of all data (including H and L). After removing H and L, the new data is the non - outlier data (between whiskers). The median of this new data: since the original median was in the middle of the non - outlier data (because the outliers are at the ends), the median of the non - outlier data is the same as the original median. For the mean: the high outlier is a large value, so (sum - H - L) compared to original sum: if H > L (which it is, since high outlier is above upper whisker, low below lower whisker, so H is much larger than L in magnitude? Wait, no – the low outlier is a small number (like 0 or 5), high outlier is a large number (like 50). So sum - H - L = (sum) - (large + small). The original mean was (\frac{sum}{n}), new mean is (\frac{sum - H - L}{n - 2}). Let's take an example: suppose data is [1, 20, 25, 30, 50] (outliers: 1 and 50, median 25). Sum = 1+20+25+30+50 = 126, mean = 126/5 = 25.2. Remove outliers: [20,25,30], sum = 75, mean = 75/3 = 25. So mean decreased? Wait, no, in this case, the high outlier (50) was increasing the mean, low outlier (1) decreasing. Removing both: sum decreases by 51 (50 + 1), n decreases by 2. Original mean 25.2, new mean 25. So mean decreased. But wait, in the problem, the high outlier is from a professional contest (very high) and low from children’s (very low). Wait, maybe my example is wrong. Wait, the box plot: the median is the middle line. When we remove outliers, the median of the remaining data (non - outlier) is the same as the original median? No, in the example, original median was 25 (middle of 5 data points). After removing 2 outliers, we have 3 data points, median is 25 (same as original). So median remains unchanged. The mean: original mean included a high outlier (increasing mean) and low outlier (decreasing mean). But the high outlier is much larger (professional contest: eats many pies) than the low outlier (children’s: eats few). Wait, no – the low outlier is a small number (like 5), high outlier is a large number (like 50). So sum - H - L: if H is much larger than L, then (sum - H - L) is less than sum - L (since H is large). So the mean, which was (\frac{sum}{n}), now is (\frac{sum - H - L}{n - 2}). Since H is a large positive value, removing it (and L, a small negative - effect value) will decrease the mean? Wait, no, in the example above, mean decreased. But wait, the options: one of the options is "The median will remain unchanged, and the mean will decrease". Wait, let's re - analyze:

Median: the median is the middle value of the data set. When we remove outliers (which are at the extremes), the middle value of the remaining data (non - outlier) is the same as the original median, because the original median was in the middle of the non - outlier data (the outliers are just at the ends, not affecting the middle position of the main data).

Mean: the mean is affected by outliers. The high outlier (large value) was increasing the mean, the low outlier (small value) was decreasing the mean. But the high outlier is a very large value (from a professional contest) and the low outlier is a very small value (from a children’s contest). So the high outlier has a larger magnitude in increasing the mean than the low outlier has in decreasing it. Wait, no – actually, when you remove both, the sum decreases by (high outlier + low outlier). The original mean is (\frac{\sum x}{n}), new mean is (\frac{\sum x - high - low}{n - 2}). Let's take a better example: suppose data is [5, 20, 25, 30, 50]. Outliers: 5 (low, since 5 < Q1 - 1.5IQR? Wait, no, let's calculate IQR. Q1 = 20, Q3 = 30, IQR = 10, 1.5IQR = 15. So Q1 - 15 = 5, so 5 is not an outlier (it's at Q1 - 1.5IQR). Oops, bad example. Let's make outliers: [1, 20, 25, 30, 60]. Q1 = 20, Q3 = 30, IQR = 10, 1.5IQR = 15. Q1 - 15 = 5, so 1 < 5 (outlier). Q3 + 15 = 45, 60 > 45 (outlier). Now, original median: 25. Sum = 1+20+25+30+60 = 136, mean = 136/5 = 27.2. Remove outliers: [20,25,30], sum = 75, mean = 75/3 = 25. Median of new data: 25 (same as original). So mean decreased (27.2→25), median unchanged. So in this case, removing both outliers (high and low) – the high outlier was increasing the mean, low outlier decreasing. But the high outlier's effect on increasing the mean was stronger than the low outlier's effect on decreasing? No, in this case, sum decreased by 1 + 60 = 61, n decreased by 2. Original mean 27.2, new mean 25. So mean decreased. The median remains the same because the middle value of the non - outlier data is the same as the original median. So the correct option should be "The median will remain unchanged, and the mean will decrease".

Answer:

The median will remain unchanged, and the mean will decrease.