Library Survey with Missing Frequencies
A school library survey records the number of pages read in a week by 50 borrowers. The librarian has the class intervals 0–10, 10–20, 20–30, 30–40 and 40–50, with frequencies 5, f1, 15, f2 and 8, but the two entries f1 and f2 were smudged. She remembers that the total number of borrowers is 50 and that the median number of pages read is 28. She also wants to test a claim that 'the average borrower read more than 30 pages in the week'. Use the median formula for grouped data and the mean formula for grouped data, taking class marks at the midpoints. Assume the data are continuous and that the median lies in the 20–30 class. The librarian needs both missing frequencies before she can compute the mean, and she wants the claim judged on the computed mean, not on the median.
1. Find the missing frequencies f1 and f2.
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Total frequency: 5 + f1 + 15 + f2 + 8 = 50, so f1 + f2 = 22.
Median = 28 lies in 20–30. For this class, l = 20, h = 10, f = 15. Cumulative frequency before the class = 5 + f1. N/2 = 25.
Median formula: 28 = 20 + [(25 − (5 + f1))/15] × 10.
So 8 = (10/15)(20 − f1) = (2/3)(20 − f1). Thus 24 = 40 − 2f1, giving 2f1 = 16 and f1 = 8. Then f2 = 22 − 8 = 14.
2. Compute the mean number of pages read.
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Class marks: 5, 15, 25, 35, 45. Products: 5×5=25; 8×15=120; 15×25=375; 14×35=490; 8×45=360. Sum = 25+120+375+490+360 = 1,370.
Mean = 1370/50 = 27.4 pages.
3. Judge whether the claim that the average borrower read more than 30 pages is supported.
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The claim is that the average borrower read more than 30 pages. The computed mean is 27.4 pages, which is less than 30. Therefore the claim is not supported by the data. The median being 28 also does not support an average above 30; it suggests the middle borrower read fewer than 30 pages.