The first half, saying the mean scores are the same, is wrong. Both classes do share the same mode, a score of 5 (9 students), but the frequencies lean to different sides: in Class T, 16 students fall above 5 points (scores 6 to 9), while in Class U, 16 students fall below (scores 1 to 4). The mean scores are therefore about 5.77 and about 4.23 points, a difference of roughly 1.53 points. Matching modes do not guarantee matching means, and that is the trap in this option. The second half, saying the standard deviations differ, is also wrong: both are about 1.45.
○イ
正解Reading the frequencies from Figure 4: Class T has 2 students scoring 3, 3 scoring 4, 9 scoring 5, 7 scoring 6, 5 scoring 7, 3 scoring 8, and 1 scoring 9 (30 students in total); Class U has 1 student scoring 1, 3 scoring 2, 5 scoring 3, 7 scoring 4, 9 scoring 5, 3 scoring 6, and 2 scoring 7 (30 students in total). The total for Class T is 3×2+4×3+5×9+6×7+7×5+8×3+9×1 = 173, so its mean score is 173÷30 ≒ 5.77 points; the total for Class U is 1×1+2×3+3×5+4×7+5×9+6×3+7×2 = 127, so its mean score is 127÷30 ≒ 4.23 points. The mean scores are therefore different. On the other hand, looking at the sequence of frequencies, Class U is exactly the shape of Class T mirrored about a score of 5 (the number of students scoring k in Class U equals the number scoring 10−k in Class T), so the spread of the distribution is identical. Standard deviation is a measure of how far the data lie from the mean, so its value does not change when a distribution is shifted or mirrored. Calculating it confirms this: both classes have a variance of ≒ 2.11 and a standard deviation of ≒ 1.45. Therefore イ, which states that the mean scores are different but the standard deviations are the same, is the correct answer.
×Incorrectウ
It is true that the standard deviations are the same, but saying the mean scores are also the same is wrong. Class T is distributed over scores 3 to 9 and Class U over scores 1 to 7, so the two ends appear to be offset by 2 points; however, this is not a shift of the whole distribution but a mirroring about a score of 5. The difference in mean scores is therefore not 2 points but about 1.53 points (5.77 points versus 4.23 points). Either way, the mean scores do not match.
×Incorrectエ
It is true that the mean scores are different, but saying the standard deviations are also different is wrong. The distribution of Class U is simply the distribution of Class T mirrored horizontally, so the breakdown of how far each data point lies from the mean is identical in both classes. Calculation confirms that both have a variance of about 2.11 and a standard deviation of about 1.45 points, an exact match. Even if the position of the distribution is offset, the standard deviation does not change as long as the shape is the same.
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The mean expresses the position (center) of a distribution while the standard deviation expresses its dispersion (spread); they are separate measures, and one being the same does not mean the other is too. Note also that matching modes do not imply matching means.
If a frequency distribution is shifted or mirrored horizontally, the standard deviation (and the variance) stays the same as long as the shape is unchanged; only the mean changes.