# General Report on Tunny

23d Page 82

 BR 3407 0450/24/2 WD 23/2 COL 7 T 6020 1P2/L R 3106 A 1553 Φ 27.8 ST 1624 K1 K2 02 08 E 1631 02 20 E 1625 11 09 A 1633 16 07 A 1629 18 08 D 1629 19 31 C 1625 37 21 E 1637 40 28 B 1655    B 102  3.7Φ 40 28 C 1655 SET K1 K2  40 28 SPAN IN 1000'S 0271 0245 0268 0249 0301 0214 0280 0237 0268 0247 0261 0257 SPAN 4000-END IN 500'S 0139 0119 0129 0128 0131 0128 0130 0129 SPAN 01-4500 1259 1064    2B  195 B 98   4.1Φ SPANNING 01-4500 SET K1 K2  40 28  GOOD The items in the first line are message number, time and date sent, wheel day and Colossus number.     T 6020 is the text length, measured as a check, as soon as the tape is on Colossus.     1P2/L is typewriterese for 1+2/ Χ2 x; this run is chosen because the chit (not preserved) was so marked [23E].     R 3106 is the number of places looked at, i.e. the number of places where Χ2 = x.     At random the expected number (A) of these when 1+2 = . is½ × 3106 = 1553.     Φ (typewriterese for σ) 27.8 is the standard deviation of 1+2 = . viz. ½ = ½ . This is of course an application of the formula quoted in 21 (b), that if random proportional frequency in a normal distribution is p, the standard deviation is . A table of ½ and ¼ is provided at each Colossus.     ST 1624 is the set total, i.e. Colossus is set so as not to display or print any smaller score. Because this is a two-wheel run, ST is taken as A + 2½σ.     The best score is 3.7σ, not even "good" [23C(a)] but worth spanning [23F(o)]. In each pair of span scores the upper is 1+2 = ., the lower 1+2 = x : this makes it easy to see where a slide occurs, evidently between 4000 and 5000, for 5000 - 6000 shows almost no bulge and 4000 - 5000 only a small bulge. 4000 - 5000 is therefore spanned in 500's and the bulge of 1+2 = . is seen to cease at about the 4500th letter: it is therefore believed that there is a message slide here and the subsequent runs are done spanning 1 - 4500: the sigma-age is now 4.1 instead of 3.7; the setting is therefore "good".     Here the operator makes the mistake of neglecting a 4-letter count for ΔD1, ΔD2 [23E(h)] this is easily reconstructed and would read         .   .    592         .  x    498         x x    667         x  .    567 Because xx is so strong, this would have suggested the run 3+4x/1x2x which would in fact yield a score of 7.9σ.