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Exercise 10.2 · Q9

Q.Calculate reasonable value for PR70PR_{70} and PR80PR_{80} for example no. 43 (given in Percentile rank of grouped data).

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The method for PR70PR_{70} and PR80PR_{80} (the 70th and 80th percentiles of grouped data) follows directly from the percentile formula applied twice with k=70k=70 and k=80k=80 — but this question explicitly reuses Example 43's frequency table from the textbook, which is not part of the extracted content available to us, so we present the exact method rather than invent class-interval figures.

For a grouped (continuous) frequency distribution with total frequency N=∑fN=\sum f:

Pk=l+kN100−cff×hP_k = l + \frac{\dfrac{kN}{100}-cf}{f}\times h

where ll = lower boundary of the class containing PkP_k (found by locating kN100\frac{kN}{100} in the cumulative-frequency column), cfcf = cumulative frequency of the class just before it, ff = frequency of the PkP_k-class, hh = class width.

  1. From Example 43's cumulative-frequency (cfcf) column, compute the two target ranks: 70N100=0.70N\dfrac{70N}{100}=0.70N and 80N100=0.80N\dfrac{80N}{100}=0.80N.
  2. Locate the class-interval in which each rank falls: the class whose cumulative frequency first equals or exceeds 0.70N0.70N contains PR70PR_{70}; similarly for 0.80N0.80N and PR80PR_{80}.
  3. For PR70PR_{70}: read off that class's lower boundary ll, the cumulative frequency cfcf of the class immediately before it, its own frequency ff, and the class width hh; substitute into P70=l+0.70N−cff×hP_{70}=l+\dfrac{0.70N-cf}{f}\times h.
  4. For PR80PR_{80}: repeat step 3 with the class containing 0.80N0.80N, substituting into P80=l+0.80N−cff×hP_{80}=l+\dfrac{0.80N-cf}{f}\times h.
  5. Since 80%>70%80\% > 70\%, expect P80>P70P_{80} > P_{70} as a self-check (a higher percentile rank must correspond to a value further up the distribution) — if the arithmetic gives P80≤P70P_{80}\le P_{70}, re-check the class-location step. …

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