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Question
as constructed for a data set, but it was later discovered that the maximum value was recorded incorrectly. t been 94.
uch does the range of the data increase after making this correction?
se in range = 94 − 53
se in range = 41
eat effort
making this correction, which quartile shows the greatest spread of data?
first quartile (q₁)
second quartile (q₂)
third quartile (q₃)
fourth quartile (q₄)
First Sub - Question (Increase in Range)
Step 1: Recall the formula for range
The range of a data set is calculated as \( \text{Range}=\text{Maximum}-\text{Minimum} \). Let the original maximum be \( M_1 \) and the corrected maximum be \( M_2 = 94 \). The increase in range will be the difference between the new range and the old range. Since the minimum value does not change, the increase in range is \( M_2 - M_1 \). From the given calculation, \( M_1 = 53 \) (implied from \( 94 - 53 \)).
Step 2: Calculate the increase in range
We calculate \( 94-53 \).
\( 94 - 53=41 \)
Quartiles divide a data set into four equal parts. The first quartile (\( Q_1 \)) is the median of the lower half, the second quartile (\( Q_2 \)) is the overall median, and the third quartile (\( Q_3 \)) is the median of the upper half. When the maximum value (which is in the upper half of the data set) is corrected to a larger value (94), the upper half of the data set (from \( Q_2 \) to \( \text{Maximum} \)) will have a greater spread. The third quartile (\( Q_3 \)) is part of the upper half, and the spread related to \( Q_3 \) (the distance from \( Q_2 \) to \( Q_3 \) and from \( Q_3 \) to the new maximum) will be affected. Since the maximum is increased, the upper half (where \( Q_3 \) lies) will show the greatest spread. Also, note that there is no fourth quartile in the standard quartile system (quartiles are \( Q_1, Q_2, Q_3 \) dividing the data into four parts). So among the given options, the third quartile (\( Q_3 \)) shows the greatest spread after the correction.
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The range of the data increases by 41.