Exercise 23.4

*Q1) 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. *

Solution:

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.

Median = Value of \(\frac{n+1}{2}th\)^{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 â€“ 2 Mean

=> 14 = 3 x 14 â€“ 2 Mean

=> 2 Mean = 42 â€“ 14 = 28

=> Mean = 28 \(\div\)

*Q2) Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34 *

Solution:

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

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 35.

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

*Q3) Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4 *

Solution:

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.

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*Q4) 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. *

Solution:

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

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

Here, n = 11

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 15.

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

Now,

Mode = 3 Median â€“ 2 Mean

=> 10 = 3 x 15 â€“ 2 Mean

=> 2 Mean = 45 â€“ 10 = 35

=> Mean = 35 \(\div\)

*Q5) Find the mode of the following data: *

*12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14 *

Solution:

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, 18

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

Therefore, 14 is the mode of the given data.

*Q6) 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, 164, 163, 160, 165, 163, 162 *

*What is the mode of heights? *

*Also, find the mean and median. *

Solution:

Arranging the data in tabular form, we get:

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

160 | lll | 3 |

161 | l | 1 |

162 | llll | 4 |

163 | llll llll | 10 |

164 | lll | 3 |

165 | lll | 3 |

168 | l | 1 |

Total | 25 |

Here, n = 25

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 163 cm.

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

Mode = 3 Median â€“ 2 Mean

=> 163 = 3 x 163 â€“ 2 Mean

=> 2 Mean = 326

=> Mean = 163 cm.

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*Q7) 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? *

Solution:

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

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 20.

Here, clearly, 20 occurs 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.

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*Q8) Calculate the mean and median for the following data: *

*Marks: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10 Â Â Â Â Â Â 11 Â Â Â Â Â Â 12 Â Â Â Â Â Â 13 Â Â Â Â Â Â 14 Â Â Â Â Â Â 16Â Â Â Â Â Â Â 19Â Â Â Â Â Â Â 20*

*Number of students: Â Â 3 Â Â Â Â Â Â Â Â 5 Â Â Â Â Â Â Â Â 4 Â Â Â Â Â Â Â Â 5 Â Â Â Â Â Â Â Â 2 Â Â Â Â Â Â Â Â 3 Â Â Â Â Â Â Â Â 2Â Â Â Â Â Â Â Â Â 1*

*Using empirical formula, find its mode. *

Solution:

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 | \(\sum f_{i}\) |

\(f_{i}x_{i}\) |
30 | 55 | 48 | 65 | 28 | 48 | 38 | 20 | \(\sum f_{i}x_{i}\) |

Mean = \(\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{332}{25}\)

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

Median = Value of \(\frac{n+1}{2}th\)^{th} observation = 13.

Now,

Mode = 3Median â€“ 2 Mean

=> Mode = 3 (13) â€“ 2 (13.28)

=> Mode = 39 â€“ 26.56

=> Mode = 12.44.

*Q9) The following table shows the weights of 12 persons. *

*Weight (in kg): Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 48 Â Â Â Â Â Â 50 Â Â Â Â Â Â 52 Â Â Â Â Â Â 54 Â Â Â Â Â Â 58 *

*Number of persons: Â Â Â Â Â Â Â Â Â Â Â Â Â 4 Â Â Â Â Â Â Â Â 3 Â Â Â Â Â Â Â Â 2 Â Â Â Â Â Â Â Â 2 Â Â Â Â Â Â Â Â 1 *

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

Solution:

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

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

\(f_{i}x_{i}\) |
192 | 150 | 104 | 108 | 58 | \(\sum f_{i}x_{i}\) |

Mean = \(\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{612}{12}\)

Here, n = 12

Median = \(\frac{n}{2}th\)

=> Median = \(\frac{Value\;of\;6^{th}\;observation+Value\;of\;7^{th}\;observation}{2}\)

=> Median = \(\frac{50+50}{2}\)

Now,

Mode = 3 Median â€“ 2 Mean

=> Mode = 3 x 50 â€“ 2 x 51

=> Mode = 150 â€“ 102

=> Mode = 48 kg.

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