Question

1. Given below are marks obtained by 20 students out of 25 in Mathematics test. 21 23 19 17 12 15 15 17 17 19 23 23 21 21 25 25 21 19 19 19 Calculate: i. measures of central tendency. (9 marks) ii. range and semi inter-quartile range. (6 marks) iii. variance and determine the shape of skewness. (5 marks)

Answer 1

1) Page No: cis measures of Central tendency: Given data : 23 21, 23, 19, 17, 12, 15, 15, 17, 17 19 , 23, 25, 25.,21 21, 21, 19, 19, lg Here total numbers=n=20 - mean = [217 23 + 19 +17+12+15+15+ 17+17+19+ 23 +23+25+25+21+ 2) + 19 +19 +19+21) 20 391 - mean 20 .. mean = 19.55 1 Median : arrange the observations in ascending order 12, 15, 15, 17, 17, 17, 19, 19, 19, 19, 19, 21, 21, 21, 21, 23, 23, 23,25,25 h=20 tevens Median value of C9 )"observation +(2+1)* oberration th value of (20) observation (22) th Observation + Median = value of (lajth observation + (10th Observation Median = - 2 (396.b. 2 2

Page No:2 Median = 19 +19 38 2 2 Median 19 mode! In the given data, the observation 19 occurs maximum number of times (5 times) Mode - - 1 g (i) Range e- - Largest observation - smallest observation - 25 12 Range = 13 Here third quartile = Q3 = quartile = Qs = {e coun) value of the observation 3 (20+1)/ value of the observation . Q 3 = [15.75] + halue of the observation = [15]th observation + 0.75 [16th 15th] = 21 +0.75 [23–21] - 21 +0.75 [2 M

Page No :3 Q₂ = 21+ 1.5 Q3 = 22.5 first quartile=Q,= [n+1] Value of nt value of the observation 4 4. = [ 21 ] + value of the observation Q1 = [5.25]"value of the observation Q, - [5] *hobservation +0.25 5 [6* -51] 17 +0.25 [17-17 - 17 + Q - 17 Interquartile range = IQR = Q3 -Q, IQ R - 22.5 - 17 IQ R = 5.5 Semi- Interquartile range SIR = = SIR= Q3-QI 23-01 5.5 2 SIR = 2.75

Page No : 4 i Variance (Sample Variance) (5²) dx dx² x = x - 20 1 1 3 1 21 23 19 17 12 15 - 3 ها سه همی ها 25 -5 25 15 3. 9 g - 3 17 17 19 1 - 1 La 23 23 g 3 21 imm ī morannett 1 1 21 5 25 25 25 5 25 1 21 19 1 1 1 19 - 1 19 2x=391 Edx=-1 Edx² = 227 Here T= 19.55 But so We took iteger since it is not 5=20 an

Sample variance - s² = {dx² Page No: 5 (Edx)" fem n (-9)2 227 se 20 (1 19 227 - 4.05 ०२ 19 s² 11.7342 Standard deviation =s=√11.7342 s-3.4255 From histogram ,it can solo be that the shape is skewed to left

Page No : 6 b) probability of defective = 1% P = 1) = 1 10o P = 0.01 There are Here we distribution total n=10 boxes. are going to use poisson P(x=0) =ém. mm ri = np 3. - mean m=10X0.0 Im=o. i Exactly one defective component: .. Here r=1 P (8=1) 1 eot (o.s) = 0.1 ero. (o. 11 -0.1 P (r=1) - e xo.) = 0.9048 X 0.1 Pral) =0.09048

Page No:7 ii) at least two least two defective: ke p(872) - P(r=2) +P (r=3) + ... + P(r=10), -- - [P(r=0) + P(r =) ē ° (0:1) + 0.09048 0.1 0 - 1 - ol -1- Leo [ {-0.) to .ogo 48 1-[ 0.09048 o.9048 + 0.09048 POV> 2) = 0.00472 رنز) It contains between 1 and 4 P (r= 1,2,3,4) = P (r= t p (r=2+ P (r=3) + p (r= 4) 0.09048+ éo (0.1) 2 2) t é (0.1) ) 0 ) 3 (old 3) + 41,

= LI = Format Painter B Clipboard Conditional Formatting - Table Sort & Find & Filter - Select * Marks of the students in mathematics Marks of students Type here to search *) IN 2013,