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If the median of the following frequency distribution is 32.5. Find the values of f_{1} and f2.

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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#### Solution

**Given: **Median = 32.5

We prepare the cumulative frequency table, as given below.

Class interval: |
Frequency: (f_{i}) |
Cumulative frequency (c.f.) |

0-10 | f_{1} |
f_{1} |

10-20 | 5 | 5 + f_{1} |

20-30 | 9 | 14 + f_{1} |

30-40 | 12 | 26 +f_{1} |

40-50 | f_{2} |
26 +f_{1} + f_{2} |

50-60 | 3 | 29 + f_{1} + f_{2} |

60-70 | 2 | 31 + f_{1} + f_{2} |

N = 40 = 31 + f_{1} +f_{2} |

Now, we have

N = 40

31 + f_{1} + f_{2} = 40

f_{2} = 9 - f_{1 } ........(1)

Also, `("N")/(2) = 20`

Since median = 32.5 so the median class is 30 - 40.

Here, l = 30, f = 12 , F = 14 + f_{1} and h = 10

We know that

Median = `"l" + {{("N")/(2) -"F")/("f")} xx "h"`

`32.5 = 30 + {(20-(14+"f"_1))/(12)} xx 10`

`2.5 = ((6 - "f"_1) xx 10)/(12)`

2.5 x 12 = 60 - 10`"f"_1`

`"f"_1 = (30)/(10)`

= 3

Putting the value of `f_1` in (1), we get

`"f"_2` = 9 - 3

= 6

Hence, the missing frequencies are 3 and 6.

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