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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

#### 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. ofaccidents (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. ofaccidents (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.

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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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