Pocket money (in ₹ , per month) of 30 students is 75 70 71 55 38 52 40 48 60 65 35 58 70 61 65 64 63 58 50 53 62 64 66 65 75 64 57 60 63 45 Form a frequency table with a grouping of 30 - 40 , 40 - 50 etc. Also find the modal pocket money (by graphical method) as well as by using formula.

Pocket Money Number of students f 30 - 40 3 40 - 50 3 50 - 60 8 60 - 70 13 70 - 80 3 Σ f = 30 Since, 13 is the highest frequency, corresponds to this modal class is 60 - 70 . As we know, Mode = l + f m - f m - 1 2 f m - f m - 1 - f m + 1 × h Here, lower limit of the modal class = l = 60 , Frequency of the modal class = f m = 13 , Frequency of the class preceding the modal class = f m - 1 = 8 , Frequency of the class succeeding the modal class = f m + 1 = 3 Width of the modal class = h = 10 , Mode = 60 + ( 13 - 8 ) ( 2 × 13 - 8 - 3 ) × 10 = 60 + 5 15 × 10 = 60 + 3 · 33 Mode = 63 . 33 By graphically, We construct a relevant portion of the histogram and estimate it:- Since mode is dependent only on neighbourhood frequencies, drawing the rectangle corresponding to modal class and adjoining two classes. Joining the ends of opposite rectangles by straight lines. The point of intersection of these straight lines corresponds to mode. Two lines meet at point A , which corresponds to value of 63 . 3 ₹ Hence, by graphically, mode = 63 . 3 ₹ By formula, mode = 63 . 33 ₹

Find the mode of the following distribution by drawing a histogram Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class.

Mid value Frequency Class interval 12 20 9 - 15 18 12 15 - 21 24 8 21 - 27 30 24 27 - 33 36 16 33 - 39 42 8 39 - 45 48 12 45 - 51 It is easy to see that the highest frequency is 24 corresponding to this class interval is 27 - 33 . Hence, The modal class is 27 - 33 . We construct the relevant portion of histogram and estimate mode from it: Since mode is dependent only on neighbourhood frequencies, drawing the rectangle corresponding to modal class and adjoining two classes. Joining the ends of opposite rectangles by straight lines - the point of intersection corresponds to mode. The two lines meet at a point A , which corresponds to x - value of 31 . Hence, Mode = 31 .

Draw a histogram for the following distribution: Wt. in kg 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 Number of students 2 8 12 10 6 4 Hence, estimate the modal weight.

Wt in kg Number of students ( f ) 40 - 44 2 45 - 49 8 50 - 54 12 55 - 59 10 60 - 64 6 65 - 69 4 It is easy to see that the highest frequency is 12 corresponding to this class interval is 50 - 54 . The modal class is 50 - 54 Now constructing a histogram of given data: Since mode is dependent only on neighbourhood frequencies, drawing the rectangle corresponding to modal class and adjoining two classes. Joining the ends of opposite rectangles by straight lines - the point of intersection corresponds to mode. The Two lines meet at point A , which corresponds to value of 53 . Mode = 53 kg .

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What is no mode condition?

Important Questions on Statistics

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Class:	$0 - 6$	$6 - 12$	$12 - 18$	$18 - 24$	$24 - 30$
Frequency:	$a$	$b$	$12$	$9$	$5$

class	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
Frequency	$α$	$110$	$54$	$30$	$β$

Class :	$3 - 6$	$6 - 9$	$9 - 12$	$12 - 15$	$15 - 18$	$18 - 21$	$21 - 24$
Frequency	$2$	$5$	$10$	$23$	$21$	$12$	$3$

Class :	$0 - 7$	$7 - 14$	$14 - 21$	$21 - 28$	$28 - 35$	$35 - 42$	$42 - 49$
Frequency	$19$	$25$	$36$	$72$	$51$	$43$	$28$

$75$	$70$	$71$	$55$	$38$	$52$	$40$	$48$	$60$	$65$
$35$	$58$	$70$	$61$	$65$	$64$	$63$	$58$	$50$	$53$
$62$	$64$	$66$	$65$	$75$	$64$	$57$	$60$	$63$	$45$

Mid value	$12$	$18$	$24$	$30$	$36$	$42$	$48$
Frequency	$20$	$12$	$8$	$24$	$16$	$8$	$12$

Wt. in $kg$	$40 - 44$	$45 - 49$	$50 - 54$	$55 - 59$	$60 - 64$	$65 - 69$
Number of students	$2$	$8$	$12$	$10$	$6$	$4$

What is no mode condition?

Important Questions on Statistics

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