# 7th Class Mental Ability Data Handling

Data Handling

Category : 7th Class

Data Handling

Learning Objectives

• Data Collection
• Graphical Representation of Data
• Probability

Data Collection

Data collection is important as it helps us to collect, study and record information, make decisions about important issues as well as to pass Information on to others. It enables us to provide information regarding a specific topic or area of study.

The data that are collected need to be organised in a proper table, so that it becomes easy to understand and interpret.

Lets understand some important terms:

• Frequency: The frequency of a particular data value is the number of times the data value occurs. For example, if four students have a score of 80 in mathematics, and then the score of 80 is said to have a frequency of 4. The frequency of a data value is often represented by f.
• Variable: The measurable characteristics like age, height. Income etc. are called variables.
• Mean: The mean or arithmetic mean is the average of the numbers. Average is a number that represents or shows the central tendency of a group of observations or data. It can be calculated by using the formula given below:

Mean =$\frac{\sum{fixi}}{\sum{fi}}$. (where f is frequency and x is variable).

• Mode: It is another form of central tendency or representative value. The mode in a list of numbers refers to the list of numbers that occur most frequently.

For example, to find the mode of 9, 11, 3, 44, 17, 11, 17, 15, 15, 15, 27, 40, 8

Put the numbers in increasing order as shown below:

3, 8, 9, 15, 15, 15, 17, 17, 17, 17, 27, 40, 44

The number which occur most frequently is the mode. In the given data, the mode is 17 which occurs the most at 4 times.

• The median: The Median is the ‘middle value’ in your list. It is also a form of representative value. When the totals of the list are odd, the median is the middle entry in the list after sorting the list into increasing order If a set of n observations is given, where n is odd, then

Median = ${{\left( \frac{n+1}{2} \right)}^{th}}$observation

When observations are arranged in an increasing order.

When the totals of the list are even, the median is equal to the sum of the two middle (after sorting the list into increasing order) numbers divided by two.

If a set of n observations is given, where n is even, then

Median = $\frac{{{\left( \frac{n}{2} \right)}^{th}}observation+{{\left( \frac{n}{2}+1 \right)}^{th}}observation}{2}$

When observations are arranged in increasing order.

• Range: The difference between the smallest and largest observation is called the range.

Example

The mean, median, mode and range for the following list of values are:

8, 9, 10, 10, 10, 11, 11, 11, 11, 13

(a) 10.5, 10.5, 10, 11      (b) 10.5, 10.4, 10, 5

(c) 10.4, 10.5, 11, 5        (d) 10.5, 10.4, 11, 11

(e) None of these

Ans.     (c)

Explanation: The mean is the usual average.

$\therefore \,\,\,\,Mean\,\,\,=\,\,\,\frac{\left( 8\text{ }+\text{ }9\text{ }+\text{ }10\text{ }+\text{ }10\text{ }+\text{ }10\text{ }+\text{ }11\text{ }+\text{ }11\text{ }+\text{ }11\text{ }+\text{ }11\text{ }+\text{ }13 \right)}{10}\,\,\,\,=\,\,\,\frac{104}{10}\,\,\,=\,\,10.4$

The list contains 10 observations, so the median is given by

Median = $\frac{{{\left( \frac{n}{2} \right)}^{th}}observation+{{\left( \frac{n}{2}+1 \right)}^{th}}observation}{2}$

$\frac{{{5}^{th}}\,observation\,\,+\,\,{{6}^{th}}\,\,observation}{2}\,\,\,=\,\,\,\frac{10+11}{2}\,\,\,=\,\,\,\frac{21}{2}\,\,\,=10.5$

The mode is the number repeated most often. In this list, 11 is repeated four times.

Thus, mode = 11.

The largest value is 13 and the smallest is 8. Thus, the range is $13 - 8 = 5$.

Graphical Representation of Data

Data can be represented as bar graph, double bar graph, pie chart etc.

• Bar graph: A bar graph (also called bar chart) is a graphical display of data using bars of different heights. The bars can be plotted vertically or horizontally.
• A bar graph having double bars help to compare two collections of data at a glance.
• Pie chart: It is another way of representing the given data in which each section proportionately represents the data.

The angle of a section$=\frac{data\,corresponding\,to\,the\,\sec tion}{sum\,of\,data}\times 360{}^\circ =\frac{{{x}_{i}}}{\sum{{{x}_{i}}}}\times 360{}^\circ .$

The result of a survey of 20 persons about the movie they like is given below. Which bar graph represent the data correctly?

Comedy            Action               Romance           Drama               Science Fiction

4                      5                            6                       1                                   4

(a)

(b)

(c)

(d)

(e) None of these

Ans.     (a)

Probability

It is the branch of mathematics that studies the possible outcomes of given events together with the outcomes’ relative likelihoods and distributions. In general, probability is a measure of the likeliness that an event will occur.

$P\left( E \right)=\frac{n(E)}{n(S)}$ where E = expected or favourable outcomes

n(E) = Number of expected or favourable outcomes

n(S) = Number of total outcomes.

Example

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

(a) $\frac{1}{3}$           (b) $\frac{3}{4}$

(c) $\frac{7}{19}$         (d) $\frac{8}{21}$

(e) $\frac{9}{21}$

Ans.     (a)

Explanation: Total number of balls = (8 + 7 + 6 = 21).

Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue.

n(E) = 7.

$\therefore \,\,\,\,P(E)\,\,\,\,=\frac{n(E)}{n(S)}\,\,\,=\,\,\,\frac{7}{21}\,\,\,=\,\,\,\frac{1}{3}$

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

(a) $\frac{1}{2}$           (b) $\frac{2}{5}$

(c) $\frac{8}{15}$         (d) $\frac{9}{20}$

(e) None of these

Ans.     (d)

Explanation: Here, S = {1, 2, 3, 4, ....... 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}

$\therefore \,\,\,P(E)\,\,\,=\frac{n(E)}{n(S)}\,\,\,=\frac{9}{20}$.

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