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The mean, median and mode of the following data will be:

Class
0-50
50-100
100-150
150-200
200-250
250-300
300-350
Frequency
2
3
5
6
5
3
1

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Mean=169, Median=170.83, Mode=174.49

Mean=170, Median=170.83, Mode=174.49

Mean=169, Median=170.83, Mode=175.49

Mean=169, Median=171.83, Mode=175.49

answer is A.

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Detailed Solution

Given,

Class	0-50	50-100	100-150	150-200	200-250	250-300	300-350
Frequency	2	3	5	6	5	3	1

The frequency distribution table will be:

Class	Frequency, fi	Mid-value, xi	Cumulative frequency	fixi
0-50	2	25	2	50
50-100	3	75	2+3=5	225
100-150	5	125	5+5=10	625
150-200	6	175	10+6=16	1050
200-250	5	225	16+5=21	1125
250-300	3	275	21+3=24	825
300-350	1	325	24+1=25	325
	Σfi=25			Σ fixi=4225

Formula used for mean:

M e a n = ∑ f i x i ∑ f i

Substituting the values:

M e a n = ∑ f i x i ∑ f i M e a n = 422525 M e a n = 169

Formula used for median:

M e d i a n = l + h × N 2 − c f f

,
where,

l

=lower limit of the median class,

h

=size of the median class,

f

= frequency of the median class,

N

= sum of frequencies and

c f

=cumulative frequency of the class just preceding the median class.
We know that,

N 2 = 252 = 12.5

, the cumulative frequency just greater than 12.5 is 16 and the corresponding class is

150 − 200

.
Therefore, the median class is

150 − 200

∴ l = 150, h = 50, f = 6, N = 25 and c f = 10

Calculating the median, we get

Median = l + h × N 2 − c f f Median = 150 + 50 × 252 − 10 6 Median = 150 + 50 × 12.5 − 10 6 Median = 150 + 50 × 2.5 6 M e d i a n = 150 + 50 × 0.416 M e d i a n = 150 + 20.83 M e d i a n = 170.83

Formula used for mode:

M o d e = 3 M e d i a n − 2 M e a n

Substituting values, we get

M o d e = 3 M e d i a n − 2 M e a n M o d e = 3 × 170.83 − 2 × 169 M o d e = 174.49

The value of mean is 169, median is 170.83 and mode is 174.49.
Thus, the correct option is option 1.

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