Question

Balls Runs 0-20 12 20-40 13 40-60 19 60-80 16 80-100 14 100-120 15 120-140 16 140-160 14 160-180 17 180-200 11 200-220 15 220-240 15 240-260 25 260-280 29 280-300 54 TOTAL 285

FIND MEAN MEDIAN AND MODE

Answer 1

Answer :

Balls	Frequency ( Runs ) ( f )	Midpoint ( x )	Cumulative Frequency	fx
0 - 20	12	10	12	120
20 - 40	13	30	25	390
40 - 60	19	50	44	950
60 - 80	16	70	60	1120
80 - 100	14	90	74	1260
100 - 120	15	110	89	1650
120 - 140	16	130	105	2080
140 - 160	14	150	119	2100
160 - 180	17	170	136	2312
180 - 200	11	190	147	2090
200 - 220	15	210	162	3150
220 - 240	15	230	177	3450
240 - 260	25	250	202	6250
260 - 280	29	270	231	7830
280 - 300	54	290	285	15660
	n = 285			$\sum _{}$fx = 50412

1 ) Mean = $\frac{{\displaystyle \sum _{}}fx}{n}$ = $\frac{50412}{285}$ = 176.88 ( Ans )

2 ) for median , first we find $\frac{n}{2}=\frac{285}{2}$ = 142.5 ,
So we can see in Cumulative Frequency column 142.5 lies in 147 , so median group = 180 - 200
We know the formula

Median = L + $\frac{\left({\displaystyle \frac{n}{2}}\right)-c{f}_b}{f_m}\times w$

Where

L is the lower class boundary of the group containing the median ( here L = 180 )
n is the total number of data , (here n = 285 )
cf_b is the cumulative frequency of the groups before the median group . ( Here cf_b = 136 )
f_m is the frequency of the median group . ( here f_m = 11 )
w is the group width , ( Here w = 20 )
So,
Median = 180 + $\frac{\left({\displaystyle \frac{285}{2}}\right)-136}{11}\times 20$

Median = 180 + $\frac{142.5-136}{11}\times 20$

Median = 180 + $\frac{130}{11}$

Median = 180 + $\frac{130}{11}$

Median = $\frac{2110}{11}$

Median = ​191.81 ( Ans )

3 ) First we find modal class , We can easily identify the modal group (the group with the highest frequency) ,
So
Modal class = 280 - 300
We know
Formula for mode

Z_​ = L_{1 + $\frac{f_1-f_0}{2f_1-f_0-f_2}$} ( L₂ - L₁ )

Where
L₁ = Lower limit of the model class . ( Here L₁ = 280 )
L₂ = Upper limit of the model class . ( Here L₂ = 300 )
f₁ = Frequency of the model class . ( Here f₁ = 54 )
f₀ = Frequency of the pre-model class . ( Here f₀ = 29 )
f₂ = Frequency of the class succeeding to the model class . ( Here f₂ = 0 )

So
Z_​ =280 _{+ $\frac{54-29}{2\left(54\right)-29-0}$} ( 300 - 280 )

Z_​ =280 _{+ $\frac{25}{108-29}$} ( 20 )

Z_​ =280 _{+ $\frac{500}{79}$}

Z_​ = <sub>$\frac{22620}{79}$

Z =</sub> 286.329 ( Ans )