CBSE Class 10 Maths Chapter 13: Statistics and Probability NCERT Solutions

1. The formula $\backslash bar\{x\}\; =\; a\; +\; \backslash frac\{\backslash sum\; f\_i\; d\_i\}\{\backslash sum\; f\_i\}$ is used for calculating the mean of grouped data using the assumed mean method. In this method, $d\_i$ represents the deviation of the class mark (mid-point) from the assumed mean 'a'. Therefore, $d\_i\; =\; x\_i\; -\; a$, where $x\_i$ is the mid-point of the class interval. The correct option is (c).

When calculating the mean of grouped data, we simplify the process by assuming that all the observations within a particular class interval are concentrated at the class mark (mid-point) of that interval. This assumption allows us to treat the data as discrete, with frequencies corresponding to these class marks. Thus, the frequencies are considered to be centered at the class marks of the classes. The correct option is (b).

Let n = \Sigma f_i be the total number of observations. The formula for the mean \bar{x} of grouped data is given by:

\bar{x} = \frac{\sum f_i x_i}{n}

From this, we can write \sum f_i x_i = n\bar{x}.

Now, consider the expression \Sigma(f_i x_i - \bar{x}). We can split the summation:

\Sigma (f_i x_i - \bar{x}) = \sum f_i x_i - \sum \bar{x}

Since \bar{x} is a constant value (the mean), the summation \sum \bar{x} means adding \bar{x} to itself n times (where n = \Sigma f_i is the total number of terms). So, \sum \bar{x} = n\bar{x}.

Substituting this back into the equation:

\Sigma (f_i x_i - \bar{x}) = n\bar{x} - n\bar{x} = 0 Therefore, \Sigma(f_i x_i - \bar{x}) is equal to 0. The correct option is (a).

The formula $\backslash bar\{x\}\; =\; a\; +\; h\backslash left(\backslash frac\{\backslash sum\; f\_i\; u\_i\}\{\backslash sum\; f\_i\}\backslash right)$ is known as the step-deviation method for calculating the mean of grouped data. In this method:

• $a$ is the assumed mean.
• $h$ is the class width (the difference between the upper and lower limits of a class interval).
• $x\_i$ is the class mark (mid-point) of the class interval.
• $u\_i$ is defined as the deviation of the class mark from the assumed mean, divided by the class width.

So, the relationship is $u\_i\; =\; \backslash frac\{x\_i\; -\; a\}\{h\}$.

This transformation simplifies calculations, especially when class widths are uniform and deviations are large.

The correct option is (c).

Cumulative frequency curves, also known as ogives, are graphical representations used to determine statistical measures like the median. Specifically, when you plot both the 'less than' ogive and the 'more than' ogive on the same set of axes, their point of intersection is significant. The x-coordinate (abscissa) of this intersection point corresponds to the median of the grouped data. The y-coordinate represents the cumulative frequency at the median. Therefore, the intersection point of the less than and more than ogives gives the median on the abscissa. The correct option is (b).

First, let's identify the median class and the modal class from the given frequency distribution.

1. Finding the Median Class:

The total frequency (n) is $10\; +\; 15\; +\; 12\; +\; 9\; +\; 20\; =\; 66$.

The position of the median is $\backslash frac\{n\}\{2\}\; =\; \backslash frac\{66\}\{2\}\; =\; 33$rd observation.

Let's calculate the cumulative frequencies (CF):

• Class 0-5: Frequency = 10, CF = 10
• Class 5-10: Frequency = 15, CF = 10 + 15 = 25
• Class 10-15: Frequency = 12, CF = 25 + 12 = 37
• Class 15-20: Frequency = 9, CF = 37 + 9 = 46
• Class 20-25: Frequency = 20, CF = 46 + 20 = 66

The cumulative frequency just greater than or equal to 33 is 37, which corresponds to the class interval 10-15. Therefore, the median class is 10-15.

The lower limit of the median class is 10.

2. Finding the Modal Class:

The modal class is the class interval with the highest frequency. In the given distribution, the highest frequency is 20, which corresponds to the class interval 20-25.

Therefore, the modal class is 20-25.

The lower limit of the modal class is 20.

3. Sum of Lower Limits:

Sum of the lower limits of the median class and the modal class = (Lower limit of median class) + (Lower limit of modal class)

= 10 + 20 = 30.

The correct option is (c).