Question

• The frequency distribution of annual salaries are given in the following table. Find; Х 0 - 20 20 - 40 40 - 60 60 - 80 5 65 25 5 a) The Mode b) What is the minimum salary of the highest % 85 ? c) What is the minimum salary of the 73. rd person? d) Draw the Box Plot For the frequency distribution given below. Find; • Mean Absolute Deviation Standard Deviation Draw the histogram 1- 32 3 - 5 8 5-7 / 7 7-9

Answer 1

1. +. Cfi °~20 5 5 6 70 25 5 95 100 60 - 80 Mode - Here maximum bequency es os coresponding class is 20.40 . Modal class = 20-40 Mo : + fito erica ^ 2 fi- fo - fix EU fi= frequency of modal class = 65 of pre-modal class = 5 to = frequency frequency of post-modal class = 20 fa = L = class = 20 Lower limit of modal width = 20 h = 5.20 + (* 20) - 20 t Too x 20 Mode = 32

Minimum salary of the highest 85% es nothing but Ps. is do -Cf Pis : L + Pis = 15th percentile Evalue of (is more) "obs. th In above 1 lth - Қаңа о = value of (is * Atos = value of 15th obs. The is cumulative frequency just 70 g corresponding class greater than 15 į 20-40 ei perc :. 1. 20 cf = c.f of class previous to percentile class f = 65 , h = 20 115 5 = 20 + 1 = 65 :: Pis lis = 23.0769

- minimum salary of the 730d person is P23 73 Page value of ( 13 )obs. • value of (73) *h obs. 13 - Cof salle w than 73 s 95 g corresponding c.f. just class i greater 40-60. 1:40, cf. = 70, f. 25, h=20 1.1: 40 6**** *2) : 123 = 40 73 - 70 - + 1 x 20 25 = 40 + 2.4 Prz 42.4

To plot Box pv4 we Knd. xa, 9. , Md, 9.X i.e. 5 number summery X • ° , Xan) : 8 - 9. - rame of mob. s value of Com)" ods = value of 25th obs. į 70 & corresponding cit just et areater greater than than as class į 20-40. lē - Cof :: Siz + | 25 - 5 20 X 20 Si26.1538 9g = vame of (3 h)* obs. = value of 45th obs c.f. just greater than 75 is gi ļ corresponding class is 40 60

+ 4+ ("5;**) = 40 + 1 - x 20 - - 25 9, = 424 44 Ma = value of () Th 665 value of (le) SDS. = value of 50th obs c.f. just greader than so i to § corresponding نو ربعك .ها- هو ... Mazitachan) + 20+ ( So 50-5 20+ / 165 - xo0) X 20 = 20 + 13.8462 = 33,8462

XC1) - 0 9. - 26. 1538 mo - 33.8662 D 93 = 44 Xcn) : 8+ Box Plot XCI) gi mo 0 10 20 30 40 50 60 70 80

filmi-ul 6.2 x 1-3 3-5 5-7 7-9 fi 2 8 7 3 mi 2 4 6 8 fimi 4 32 42 24 I mi-a 3.1 1.1 0.9 2.9 8.8 6.3 8.7 Total 102 30 mean = e = - r = Efimi 2 102 20 Mean absolute deviation = M.D. I • Efi xi- Efi Efil mi - ell the fi 30 20 = 1.5 Standard deviation I fi caci - a)2 Eti I E fi cmice)! Eti

- fim; ? 8 9 3 4 6 8 128 252 192 {fimi Efi 580 580 - (51) . / 2.99 S.D. = 1.7292 Histogram Ef i 3 5 7 9
here we find absolute mean deviation about mean. Instead of mean we may use median or mode.