**CBSE & NCERT Solutions For Class 10 Maths Statistics Solutions Chapter 14**

CBSE & NCERT Class 10 Maths Chapter 14 Statistics Solutions : In my last article for * 10 Class Maths *, I have discussed Mean and the defined ways to find Mean. In this article, we will discuss Median, Mode and

*CBSE & NCERT Online Solutions*.

__Mode__

__Mode__

Let’s define mode, the value which has the maximum frequency, or the maximum observation is known as Mode. It can also be defined as the most occurring value is known as Mode.

**Mode=I + (fi – fo) /(2fi–fo–f2) *h**

Where l = lower limit of the modal class,

h = size of the class interval

f1 = frequency of the mode class

f0 = frequency of preceding mode class

f2 = frequency of succeeding mode class.

So, we will explain few ‘*CBSE NCERT Online Solutions’ *to explain the above formula. Let’s discuss Q2 of exercise 14.2

**Q2-The following data give the information on the observed lifetimes (in hours) of 225**

Electrical components:

Lifetimes (in hours) | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |

Frequency | 10 | 35 | 52 | 61 | 38 | 29 |

Determine the modal lifetimes of the components.

Solution- The class 60-80 has the highest frequency hence

l=60

h=20

fi=61

fo=52

f2=38

Mode

Mode=l + (fi-fo/2f1-fo-f2)*h

=60 + (61-52 /2*61-52-38)*20

=60 + 9/32 *20

=60+5.625

= 65.625

The modal lifetimes of the components =65.625.

__Median__

__Median__

In a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. We will find a value within the class that will divide the entire observation set into two equal parts.

To find the observation class, we should calculate the cumulative frequencies of all the classes and n/2.

Now, we locate the class whose cumulative frequency is greater than n/2. This is called the median class. After this, we will find Median-

Median= l + (n/2-cf) / f *h

Where l = lower limit of the median class taken

n = number of observations in question

cf = cumulative frequency preceding median class,

f = frequency of median class,

h = class size considered

Let’s discuss one ‘*Online Maths Solutions For Class 10th Statistics’ *to understand the same. Q1 of exercise 14.3.

**Q1- The following frequency distribution gives the monthly consumption of electricity 68 consumers of a locality. Find the mean, median and mode of the below data and compare them.**

Monthly consumption (in units) | Number of consumers |

65-85 | 4 |

85-105 | 5 |

105-125 | 13 |

125-145 | 20 |

145-165 | 14 |

165-185 | 8 |

185-205 | 4 |

Solution: – We will discuss the median calculation only. We will calculate the cumulative frequency

Monthly consumption (in units) | Number of consumers | C F |

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 8 | 64 |

185-205 | 4 | 68 |

N=68 |

N=68 n/2=34

Median=l +{(n/2-cf)/f}*h

= 125{(34-22)/20}*20

= 125+12=137

So, now we have discussed all the three measures of central tendency as per ** Class 10 Maths syllabus**.

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