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Exercise 10.2 · Q7

Q.By multiplying each of the numbers 3, 6, 2, 1, 7, and 5 by 2 and then adding 5, we obtain the set 11, 17, 9, 7, 19, 15. What is the relationship between the standard deviations and the means for the two sets?

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Under a linear transform y=ax+by=ax+b, the mean transforms the same way (yˉ=axˉ+b\bar y=a\bar x+b) but the standard deviation only scales by ∣a∣|a| — the added constant bb shifts every value equally and cancels out of the spread.

If yi=axi+by_i = a x_i + b for constants a,ba,b, then

yˉ=axˉ+b,σy=∣a∣ σx\bar y = a\bar x + b, \qquad \sigma_y = |a|\,\sigma_x

(adding bb shifts the whole data set, so deviations from the mean — and hence σ\sigma — are unchanged; multiplying by aa scales every deviation by aa.)

  1. Original data (from Q67792): 3,6,2,1,7,53,6,2,1,7,5, with xˉ=4\bar x = 4 and σx=28/6≈2.160\sigma_x = \sqrt{28/6}\approx 2.160.
  2. Transform: yi=2xi+5y_i = 2x_i + 5, giving 11,17,9,7,19,1511, 17, 9, 7, 19, 15 (checked: 2(3)+5=112(3)+5=11, 2(6)+5=172(6)+5=17, 2(2)+5=92(2)+5=9, 2(1)+5=72(1)+5=7, 2(7)+5=192(7)+5=19, 2(5)+5=152(5)+5=15 ✓).
  3. Predicted new mean: yˉ=2(4)+5=13\bar y = 2(4)+5 = 13. Direct check: 11+17+9+7+19+15=7811+17+9+7+19+15=78, 78/6=1378/6=13 ✓.
  4. Predicted new SD: σy=∣2∣×2.160=4.320\sigma_y = |2|\times 2.160 = 4.320. …

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