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Find the Missing Frequencies and the Median for the Following Distribution If the Mean is 1.46. - CBSE Class 10 - Mathematics

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Question

Find the missing frequencies and the median for the following distribution if the mean is 1.46.

No. of accidents: 0 1 2 3 4 5 Total
Frequency (No. of days): 46 ? ? 25 10 5 200

Solution

No. of
accidents (x)
No. of days (f) fx
0 45 0
1 x x
2 y 2y
3 25 75
4 10 40
5 5 25
  N = 200 `sumf_1x_1=x+2y+140`

Given, N = 200

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ 86 + x + y = 200

⇒ x + y = 200 - 86

⇒ x+ y = 114                   ..............(1)

And Mean = 1.46

`rArr(sumfx)/N=1.46`

`rArr(x+2y+140)/200=1.46`

⇒ x + 2y + 140 = 1.46 x 200

⇒ x + 2y + 140 = 292

⇒ x + 2y = 292 - 140

⇒ x + 2y = 152                 ................(2)

Subtract equation (1) from equation (2)

⇒ x + 2y - (x+ y) = 152 - 114

⇒ x + 2y - x - y = 38

⇒ y = 38

Put the value of y in (1),

⇒ x+ y = 114

⇒ x + 38 = 114

⇒ x = 114 - 38

⇒ x = 76

Hence, the missing frequencies are 38 and 76.

(2) Calculation of median.

No. of
accidents (x)
No. of days (f) Cumulative Frequency
0 45 46
1 76 122
2 38 160
3 25 185
4 10 195
5 5 200
  N = 200  

We have

N = 200

So, `N/2=200/2=100`

Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.

Hence, the median is 1.

  Is there an error in this question or solution?

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Solution Find the Missing Frequencies and the Median for the Following Distribution If the Mean is 1.46. Concept: Median of Grouped Data.
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