Question

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Explanation:

Step1: Find the first missing value (minimum)

Range is maximum - minimum. Given maximum is 99, range = 41. So minimum = 99 - 41 = 58.

Step2: Find the third missing value (let's call it \( x \))

Q3 (third quartile) for a dataset with \( n \) values: if \( n \) is odd, Q3 is the median of the upper half. Here, let's count the number of values. Let the missing values be 58 (first), \( y \) (second), \( x \) (third). So total values: 9 (since 58, 69, 75, 78, 80, \( y \), 91, \( x \), 99). The upper half (values above the median, which is the 5th value: 80) are \( y \), 91, \( x \), 99. Wait, no, for \( n = 9 \), positions are 1 - 9. Median is position 5 (80). Upper half is positions 6 - 9: \( y \), 91, \( x \), 99. Q3 is the median of these four? Wait, no, for \( n = 9 \), Q3 is the value at position \( \frac{3(n + 1)}{4} = \frac{3(10)}{4} = 7.5 \), so average of position 7 and 8. Position 7: 91, position 8: \( x \). So Q3 = \( \frac{91 + x}{2} = 94 \). Solve: \( 91 + x = 188 \) → \( x = 97 \).

Step3: Find the second missing value (\( y \)) using mean

Mean = 81, total sum = 81 * 9 = 729. Sum of known values: 58 + 69 + 75 + 78 + 80 + 91 + 97 + 99 = let's calculate: 58 + 69 = 127; 127 + 75 = 202; 202 + 78 = 280; 280 + 80 = 360; 360 + 91 = 451; 451 + 97 = 548; 548 + 99 = 647. So \( y = 729 - 647 = 82 \).

Answer:

First missing: 58, Second missing: 82, Third missing: 97