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If the median of the following frequency distribution is 32.5, find the values of ЁЭСУ_1 тБбand тБвЁЭСУ_2. Class 0 тАУ 10 10 тАУ 20 20 тАУ 30 30 тАУ 40 40 тАУ 50 50 тАУ 60 60 тАУ 70 Total Frequency ЁЭСУ_1 5 9 12 ЁЭСУ_2 3 2 40

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If the median of the following frequency distribution is 32.5, find the values of `f_1 and f_2`.

Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Total
Frequency `f_1` 5 9 12 `f_2` 3 2 40
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Class Frequency (f) Cumulative Frequency (cf)
0 – 10 `f_1` `f_1`
10 – 20 5 `f_1` + 5
20 – 30 9 `f_1` + 14
30 – 40 12 `f_1` + 26
40 – 50 `f_2` `f_1` + `f_2` + 26
50 – 60 3 `f_1` + `f_2` + 29
60 – 70 2 `f_1` + `f_2` + 31
  N = ΣЁЭСУ = 40  

Now,` f_1 + f_2 + 31 = 40`
⇒` f_1 + f_2 = 9`
⇒ `f_2 = 9 - f_1`
The median is 32.5 which lies in 30 – 40.
Hence, median class = 30 – 40
Here, `l = 30, N/2 = 40/2 = 20, f = 12 and cf = 14 + f_1`
Now, median = 32.5
`⇒ l + ((N/2−Cf)/f) × h = 32.5`

`⇒ 30 + ((20 −(14 + f_1))/ 12)× 10 = 32.5`
`⇒ (6 − f_1)/ 12` × 10 = 2.5
`⇒ (60 − 10f_1)/12 = 2.5`
`⇒ 60 - 10f_1 = 30`
`⇒ 10f_1 = 30`
`⇒ f_1 = 3`
From equation (i), we have:
`f_2 = 9 - 3`
`⇒ f_2 = 6`

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рдЕрдзреНрдпрд╛рдп 18: Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive - EXERCISE 18╨Т [рдкреГрд╖реНрда реоренрез]

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рдЖрд░.рдПрд╕. рдЕрдЧреНрд░рд╡рд╛рд▓ Mathematics [English] Class 10
рдЕрдзреНрдпрд╛рдп 18 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive
EXERCISE 18╨Т | Q 10. | рдкреГрд╖реНрда реоренрез
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