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Statistics - Introduction

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Statistics for Grade 10 primarily focuses on calculating the measures of central tendency for grouped data: Mean, Median, and Mode.

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Mean (xˉ\bar{x}) represents the average value of the data. For grouped data, it can be calculated using the Direct Method, Assumed Mean Method, or Step-Deviation Method.

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Median represents the middle-most value of the distribution. To find the median of grouped data, we first identify the median class, which is the class where the cumulative frequency exceeds n2\frac{n}{2}.

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Mode is the value that occurs most frequently. In grouped data, the class with the highest frequency is called the modal class.

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The Empirical Relationship is a rule of thumb used to relate the three measures of central tendency: 3×Median=Mode+2×Mean3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}.

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Cumulative Frequency (cfcf) is the running total of frequencies. There are two types: 'less than' type and 'more than' type.

📐Formulae

Mean (Direct Method): xˉ=∑fixi∑fi\text{Mean (Direct Method): } \bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Mean (Assumed Mean Method): xˉ=a+∑fidi∑fi, where di=xi−a\text{Mean (Assumed Mean Method): } \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \text{ where } d_i = x_i - a

Mean (Step-Deviation Method): xˉ=a+(∑fiui∑fi)×h, where ui=xi−ah\text{Mean (Step-Deviation Method): } \bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h, \text{ where } u_i = \frac{x_i - a}{h}

Median=l+(n2−cff)×h\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h

Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h

Empirical Relation: 3 Median=Mode+2 Mean\text{Empirical Relation: } 3 \text{ Median} = \text{Mode} + 2 \text{ Mean}

💡Examples

Problem 1:

Find the mean of the following data using the Direct Method: Classes 0−100-10, 10−2010-20, 20−3020-30 with frequencies 5,8,75, 8, 7 respectively.

Solution:

First, find the class marks (xix_i): x1=0+102=5x_1 = \frac{0+10}{2} = 5, x2=10+202=15x_2 = \frac{10+20}{2} = 15, x3=20+302=25x_3 = \frac{20+30}{2} = 25. Now calculate ∑fixi\sum f_i x_i: (5×5)+(8×15)+(7×25)=25+120+175=320(5 \times 5) + (8 \times 15) + (7 \times 25) = 25 + 120 + 175 = 320 Calculate ∑fi\sum f_i: 5+8+7=205 + 8 + 7 = 20 Mean xˉ=32020=16\bar{x} = \frac{320}{20} = 16.

Explanation:

The Direct Method involves multiplying the class mark (xix_i) by its frequency (fif_i), summing them up, and dividing by the total frequency.

Problem 2:

If the mean of a distribution is 2525 and the median is 2626, find the mode using the empirical relationship.

Solution:

Using the formula: 3×Median=Mode+2×Mean3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} Substitute the given values: 3×26=Mode+2×253 \times 26 = \text{Mode} + 2 \times 25 78=Mode+5078 = \text{Mode} + 50 Mode=78−50=28\text{Mode} = 78 - 50 = 28

Explanation:

The empirical relationship allows us to find one measure of central tendency if the other two are known.

Problem 3:

Calculate the sum of frequencies for a data set where the cumulative frequency of the last class is 5050.

Solution:

The cumulative frequency of the last class interval is always equal to the total number of observations (nn). Therefore, ∑fi=50\sum f_i = 50

Explanation:

In a cumulative frequency table, the final value represents the sum of all individual frequencies in the distribution.

Introduction Class 10 Notes & Examples | CBSE Maths