The document RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Question 1:**

**Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14 By using the empirical relation also find the mean.**

Arranging the data in ascending order such that same numbers are put together, we get:

12,12,13,13, 14,14,14, 16, 19

Here*, n* = 9.

âˆ´ Median = Value of observation = Value of the 5^{th} observation = 14.

Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,

Mode = 3 Median - 2 Mean

â‡’ 14 = 3 x 14 - 2 Mean

â‡’2 Mean = 42 - 14 = 28

â‡’ Mean = 28 Ã· 2 = 14.

**Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34**

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34,35,35, 38,42.

Here*, n* = 7

âˆ´ Median = Value of observation = Value of the 4^{th} observation = 35.

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

**Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4**

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3,3,4,4,4,5,5,6.

Here, 2,3 and 4 occur three times each. Therefore, 2 ,3 and 4 are the three modes.__Alternate Solution__

Arranging the data in the form of a frequency table, we have:

Values | Tally Bars | Frequency |

0 | âˆ£ | 1 |

2 | âˆ£âˆ£âˆ£ | 3 |

3 | âˆ£âˆ£âˆ£ | 3 |

4 | âˆ£âˆ£âˆ£ | 3 |

5 | âˆ£âˆ£ | 2 |

6 | âˆ£ | 1 |

Total | 13 |

Clearly, the values 2,3 and 4 occur the maximum number of times, i.e., three times.

Hence, the mode is 2,3 and 4.

**The runs scored in a cricket match by 11 players are as follows: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10 Find the mean, mode and median of this data.**

**Answer 4:**

Arranging the data in ascending order such that same values are put together, we get:

6,8,10,10,10,15,15,50,80,100,120.

Here*, n* = 11

âˆ´ Median = Value of observation = Value of the 6^{th} observation = 15.

Here, 10 occurs three times. Therefore, 10 is the mode of the given data.

Now,

Mode = 3 Median - 2 Mean

â‡’ 10 = 3 x 15 - 2 Mean

â‡’2 Mean = 45 - 10 = 35

â‡’ Mean = 35 Ã· 2 = 17.5.

**Find the mode of the following data: 12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14**

Arranging the data in ascending order such that same values are put together, we get:

10,12,12,14,14,14,14,14,14, 16, 16, 18.

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

Alternate solution:

Arranging the data in the form of a frequency table, we get:

Values | Tally Bars | Frequency |

10 | âˆ£ | 1 |

12 | âˆ£âˆ£ | 2 |

14 | 6 | |

16 | âˆ£âˆ£ | 2 |

18 | âˆ£ | 1 |

Total | 12 |

Clearly, 14 has maximum frequency. So, the mode of the given data is 14.

**Heights of 25 children (in cm) in a school are as given below: 168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162 What is the mode of heights? Also, find the mean and median.**

Arranging the data in tabular form, we get:

Height of Children (cm) | Tally Bars | Frequency |

160 | âˆ£âˆ£âˆ£ | 3 |

161 | âˆ£ | 1 |

162 | âˆ£âˆ£âˆ£âˆ£ | 4 |

163 | 10 | |

164 | âˆ£âˆ£âˆ£ | 3 |

165 | âˆ£âˆ£âˆ£ | 3 |

168 | âˆ£ | 1 |

Total | 25 |

Here*, n* = 25

âˆ´ Median = Value of observation = Value of the 13^{th} observation = 163 cm.

Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.

Now,

Mode = 3 Median - 2 Mean

â‡’ 163 = 3 x 163 - 2 Mean

â‡’2 Mean = 326

â‡’ Mean = 326 Ã· 2 = 163 cm.

**The scores in mathematics test (out of 25) of 15 students are as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Find the mode and median of this data. Are they same?**

Arranging the data in ascending order such that same values are put together, we get:

5,9,10,12,15,16, 19, 20, 20, 20, 20, 23, 24, 25, 25.

Here*, n* = 15

âˆ´ Median = Value of observation = Value of the 8^{th} observation = 20.

Here, clearly, 20 occurs the most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.

Yes, the median and mode of the given data are the same.

**Calculate the mean and median for the folllowing data:**

Marks | : | 10 | 11 | 12 | 13 | 14 | 16 | 19 | 20 |

Number of students | : | 3 | 5 | 4 | 5 | 2 | 3 | 2 | 1 |

**Using empirical formula, find its mode.**

Calculation of Mean

Marks (x_{i}) | 10 | 11 | 12 | 13 | 14 | 16 | 19 | 20 | Total |

Number of Students (f_{i}) | 3 | 5 | 4 | 5 | 2 | 3 | 2 | 1 | |

f_{i}x_{i} | 30 | 55 | 48 | 65 | 28 | 48 | 38 | 20 |

Mean =

Here, *n* = 25, which is an odd number. Therefore,

Median = Value of observation = the 13^{th} observation = 13.

Now,

Mode = 3 Median - 2 Mean

â‡’Mode = 3 x 13 - 2 x (13.28)

â‡’Mode = 39 - 26.56

â‡’Mode = 12.44.

**The following table shows the weights of 12 persons.**

Weight (in kg): | 48 | 50 | 52 | 54 | 58 |

Number of persons: | 4 | 3 | 2 | 2 | 1 |

**Find the median and mean weights. Using empirical relation, calculate its mode.**

**Calculation of Mean**

Weight (x_{i}) | 48 | 50 | 52 | 54 | 58 | Total |

Number of Persons (f_{i}) | 4 | 3 | 2 | 2 | 1 | |

f_{i}x_{i} | 192 | 150 | 104 | 108 | 58 |

Here*, n* = 12

Now,

Mode = 3 Median - 2 Mean

â‡’ Mode = 3 x 50 - 2 x 51

â‡’Mode = 150 - 102

â‡’ Mode = 48 kg.

Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.