Question

6.92 A variable is normally distributed with mean 0 and standard deviation 4 a. Determine and interpret the quartiles of the variable. b. Obtain and interpret the second decile. c. Find the value that 15% of all possible values of the variable exceed. d. Find the two values that divide the area under the correspond- ing normal curve into a middle area of 0.80 and two outside areas of 0.10. Interpret your answer.

Answer 1

x Nero, 8-4) uartiles 9. - P ( x <$.) = 0.25 Let z = x-1 N601) P(x-4 ore) = 0.25 P(Z <0,0) = 0.25 - 01-2025 QI= -0.67 x 4 lies in lower انت 054. d all values -re value lies below 268 Q2 = Median P(Ø x 262 ) = 0.5 - P(x-e casall ) = 0.5 P(Z LQ2-0) 505 02-2015 mean Medicino

3 PCXCO3)=075 -P(x723) -0.45 PC x >03) = 0.25- P(x- e > 03-1) = 0:25 P (z 7 3 ) = 0.25 Q3 20-25 Q3 = 4x067 268 25%. & all values lies above 2.68 Quartiles are equidistant. je P(-a cx <a) = 807. P( XL-Q) = 0.10 10% a 80%. 6101 :P(X-1 (all) 2010 P(Z < 2) = 0.10 - - 70-10-7-28 - a -51 a = 5.12 је &o, s he area Ceg beaueen - 5.12 & 512

- 20/ Second elecile D2 P( x < 5) = 2 -PCX-4 < Day ) =0:2 P ( 2 (D2] = 0.2 -D2 = 70.2 = 0.94 D2 = -0.84 x 4 --3.36 below -3.36 207.89 the area lies 157 A all the values a the variable exceedsk - PC X7k) =015 P(X-4 k-1)= 0.15 P(774) = 0.15 ett = 2015 & 104 K = 4.16 0 times of the finishers in New York 1 69.5 x: the city amin) x ~Nu= 60 min, P(50 LX (70 cet 2= x 4 →N(0,1)

= P(-1.22 LALD 1 - P(Z (-1.22)-P(771) 1 E1-0-1/23 - 015866 0.73011 73.012 5) P( x 2 5 min). P(x-1 (75 ) = P(Z (75-61) = P(ZL1.56 = 1- P(2.7158) = -0.059380 = 0.94062 94.06%. 40th percentile - PLO P(x<P40) = 0.4 P(X-4 (P40-1) = 04 P(ZL P40-61) = 0.4 P40 -61 = -204 = -0.25 P40 - 9X(-0.25) +6) = 58.75 ie. 40% & the obs are below 58.75 e. 40% of the finishees required time less than 58.75 min

d) 8+decile: D8 De. P ( x (8D8 ) - 0.8 - 1-P (x > D8) -0-8 :P(X7D8) = 0.2 P(x- >D8-el) =02 - P(Z 7 D8 -1) =02 - P(77 D8-61) = 0.2 D8 - 6) = 20.2 = 0.84 9 .. Dg = 6x0.84)+6) = 68.56 20% of the finishers take time than 68.56 min. greater