# The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs Scored Number of batsmen \[3000 - 4000\] \[4\] \[4000 - 5000\] \[18\] \[5000 - 6000\] \[9\] \[6000 - 7000\] \[7\] \[7000 - 8000\] \[6\] \[8000 - 9000\] \[3\] \[9000 - 10000\] \[1\] \[10000 - 11000\] \[1\]

Find the mode of the data.

Runs Scored | Number of batsmen |

\[3000 - 4000\] | \[4\] |

\[4000 - 5000\] | \[18\] |

\[5000 - 6000\] | \[9\] |

\[6000 - 7000\] | \[7\] |

\[7000 - 8000\] | \[6\] |

\[8000 - 9000\] | \[3\] |

\[9000 - 10000\] | \[1\] |

\[10000 - 11000\] | \[1\] |

Answer

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**Hint:**In order to determine the mode of data from the distribution given in the question that shows the run scored by number of batsmen in the international cricket matches. We are given a range of data about batsmen who scored runs in a given range. The number of batsmen will be our frequency.

**Formula used:**

We can find the mode by using the formula:

\[Z = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h\].

**Complete step by step answer:**

We are given the data is a grouped continuous distribution. Runs scored are over interval and number of batsmen will be the frequency shown in the below table representation.

Runs Scored | Number of batsmen |

\[3000 - 4000\] | \[4\] |

\[4000 - 5000\] | \[18\] |

\[5000 - 6000\] | \[9\] |

\[6000 - 7000\] | \[7\] |

\[7000 - 8000\] | \[6\] |

\[8000 - 9000\] | \[3\] |

\[9000 - 10000\] | \[1\] |

\[10000 - 11000\] | \[1\] |

We will first identify the Modal Class. Modal Class is the interval with the highest frequency. In the given case, \[4000 - 5000\]has the highest frequency \[18\]so it will be the modal class.

Now we will use the formula for finding the mode,

\[Z = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h\]

\[\Rightarrow l = 4000 = \] Lower limit of the modal class

\[\Rightarrow {f_1} = 18 = \] Frequency of the modal class

\[\Rightarrow {f_0} = 4 = \] Frequency of the class preceding the modal class

\[\Rightarrow {f_2} = 9 = \] Frequency of the class succeeding the modal class

\[\Rightarrow h = 4000 - 3000 = 1000 = \]size of the class interval

Runs Scored | Number of batsmen (\[{f_i}\]) |

\[3000 - 4000\] | \[{f_0} = 4\] |

\[4000 - 5000\] | \[{f_1} = 18\] |

\[5000 - 6000\] | \[{f_2} = 9\] |

Applying the formula, we get,

\[Z = 4000 + \dfrac{{18 - 4}}{{2(18) - 4 - 9}} \times (1000)\]

\[\Rightarrow Z = 4000 + \dfrac{{14}}{{36 - 13}} \times (1000)\]

\[\Rightarrow Z = 4000 + \dfrac{{14000}}{{23}}\]

\[\Rightarrow Z = 4000 + 608.69\]

\[\therefore Z = 4608.69\] runs

**Thus, the mode of the data is \[4608.69\] runs scored.If we round off the decimal number we have 4608.7 runs scored.**

**Note:**The value that appears repeatedly in a set is known as mode. The value or number in a data set with a high frequency or appears more frequently is referred to as the mode or modal value. Apart from the mean and median, it is one of the three indicators of central inclination.In the given case, we are given continuous distribution so there is no need to make adjustments to the class intervals.

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