Data Handling

1.0Data

The word data means information in the form of numerical figures or a set of given facts. E.g. The percentage of marks scored by 10 students of a class in a test are $72,84,82,96$, 94, 98, 99, 67, 92 and 93.

Some basic definitions

Raw data: Data obtained from direct observation is called raw data. The marks obtained by 10 students in a monthly test are an example of raw data or ungrouped data.

Observation: Each numerical figure in the set of data is called an observation.

Array: Arranging the numerical figures of a set of data in ascending order is called an array. Range: The difference between the highest and lowest values of the observation in a given set of data is called its range.

Frequency: The number of times a particular observation occurs is called its frequency. The collection of a particular type of information in the form of numerical figures is called, a set of data. This set of data obtained in the original form is called a set of raw (or ungrouped) data.

Frequency distribution: The number of times a particular observation occurs is called its frequency. The table showing the frequencies of various observations of data is called a frequency distribution table or simply frequency table. We take each observation from the data and count them with the help of strokes called tally marks. For the sake of convenience, we use tally marks in bunches of five, i.e. the fifth one crossing the four diagonally.

2.0Tabulation or presentation of data

A systematic arrangement of data in a tabular form is called tabulation or presentation of the data.

While making frequency distribution table, always arrange the raw data in ascending order.

3.0Arithmetic mean

The arithmetic mean in statistics is the same as 'average' in arithmetic.

Mean of ungrouped or raw data

The mean of a set of data is found out by dividing the sum of all the observations by the total number of observations in the data. We denote the mean by $\overline{\mathrm{x}}$ (read 'x bar').

Mean $=\bar{x}=\frac{\text{ Sum of observations }}{\text{ Number of observations }}$

4.0Median

The median of a set of numbers is the middle number when all the numbers are arranged in order of size, i.e., in descending or ascending order.

Method for finding the median of an ungrouped data To find the median of a set of numbers, arrange the data in an increasing or decreasing order of magnitude. Let the total number of observations be $n$.

Case 1 : When n is odd : Median $=$ value of $\frac{1}{2}(\mathrm{n}+1{)}^{\text{th }}$ observation . Case 2 : When n is even : Then the mean of the two middle scores is the median. Median $=\frac{1}{2}\{{\left(\frac{\mathrm{n}}{2}\right)}^{\text{th }}$ observation $+{\left(\frac{\mathrm{n}}{2}+1\right)}^{\text{th }}$ observation $\}$

• Half of the values in the data set lie below the median and half lie above the median.
• While finding median of the given raw data, always arrange the data in increasing or decreasing order.

5.0Mode of ungrouped data

The mode of a set of numbers is the number which occurs most frequently in the set. If no numbers occur more than once, the set of data is said to have no mode. For example, in the following set of data $3,6,9,16,27,37,48$, no number appears more than once. Hence there is no mode available. If different numbers occur the same number of times, the set of data has more than one mode.

Note: For finding mean and mode, it is not necessary to arrange the given set of data in an ascending or descending order.

Empirical formula for calculating mode

We use the formula Mode $=3($ Median $)-2($ Mean $)$

6.0Statistical graphs

There are different types of graphs or diagrams to represent statistical data. Some of them are (i) Pictograph (ii) Bar graph (iii) Double bar graph

7.0Pictograph

It is a pictorial representation of data using symbols by observing the pictograph one can easily understand the given data. E.g., The following table gives the average hours spent by students on different activities.

we take a convenient scale to draw a pictograph for the above data

8.0Bar graph

A bar graph is a pictorial representation of numerical data in the form of rectangles (or bars) of equal width and varying heights.

• A vertical bar graph is sometimes called a column graph. These rectangles are drawn either vertically or horizontally, keeping equal space between them. The height (or length) of a rectangle depends upon the number it represents.

How to draw a bar graph: Suppose some numerical data is given to us, and we have to represent it by a bar graph on a graph paper. We can draw the graph by following the steps given below.

Step 1: On a graph paper, draw a horizontal line OX and a vertical line OY. These lines are called the x -axis and the y -axis respectively.

Step 2: Mark points at equal intervals along the x -axis. Below these points write the names of the data items whose values are to be plotted. Step 3: Choose a suitable scale. On that scale determine the heights of the bars for the given numerical values.

Step 4: Mark off these heights parallel to the $y$-axis from the points taken in Step 2.

Step 5: On the x-axis, draw bars of equal width for the heights marked in Step 4. The bars should be centred on the points marked on the x -axis. These bars represent the given numerical data.

9.0Double bar graph

A double or dual bar graph helps you to make comparisons between two related pieces of data at the same time.

• While making double bar graph, always use two colours to differentiate between the two category of data.

10.0Probability

Chance factor: There are events in our daily life which are possible or likely but uncertain. We can not be sure whether they would happen or not. E.g. When you play the play cards it is possible that you may win. You are not sure that you will win or not. Equally likely, more likely and unlikely outcomes

Equally likely

(i) If you toss a coin, you are not sure, which one, the head or tail will show up. The chances of it being either head or tail are even/equal. We say that it is equally likely that it will be either head or tail. (ii) We say that cricket is a game of luck or chance. We can not say with certainty which team will win. In a cricket match between two good teams, both teams have an even chance of winning. You may also express it by saying that it is equally likely that team A or team B will win.

Unlikely outcomes

When you buy a lottery ticket it is possible that you may win. But since lakh of people would have bought the tickets, the chances of your winning are very less. We may say that there is a very poor chance of your winning or to put it in other words it is very unlikely that you will win. Ravi is a brilliant student while Rekha is an average student. It is very likely or we may say there is a good chance that both will get promotion to next class. However, it is very unlikely, that is, there is a poor chance that Rekha will get first position in the class.

More likely outcomes

Suppose there are 16 apples and 8 bananas in a basket. Sangita wants to pick up an apple from basket. It is more likely that Sangita will pick up an apple without looking.

Some terms related to probability

Experiment: An operation which can produce some well-defined outcomes, is called an experiment. Each outcome is called an event.

Random experiment: An experiment in which all possible outcomes are known and the exact outcome can not be predicted in advance is called a random experiment.

Trial: By a trial, we mean performing a random experiment. (i) Throwing two coins : Clearly, in throwing two coins, all possible outcomes are two heads, two tails, Head on the first coin and Tail on the second, Tail on the first coin and head on the second. i.e., $\{HH,TT,HT,TH\}$ (ii) Throwing a dice : A dice is a solid cube having 6 faces, each face of which is a square. We mark these faces as $1,2,3,4,5,6$ respectively. Let us throw a dice and let it fall freely on the ground, resting on one of its faces. Then, the number on the upper face is the outcome. Thus, in throwing a dice, all possible outcomes are $1,2,3,4,5,6$.

Empirical probability

Probability uses number to measure the chance of an outcome happening. Suppose we make $n$ trials of an experiment. Then, the probability of occurrence of an event $E$ is defined as $P(E)=\frac{\text{ number of trials in which E occurs }}{\text{ total number of trials }}$

• Probability of an event is never negative.

11.0Numerical Ability

Q. Following are the ages (in years) of 10 teachers in a school. $32,41,27,54,36,25,28,57,40,38$ (i) What is the age of oldest teacher and that of the youngest teacher? (ii) Find the range of the ages of the teachers. (iii) Find the mean age. Solution: Arranging in ascending order, we get $25,27,28,32,36,38,40,41,54,57$ From the above set of data, we find that (i) Age of the oldest teacher $=57$ years Age of the youngest teacher $=25$ years (ii) Range $=(57-25)$ years $=32$ years. (iii) Mean age $(\bar{X})=\frac{\text{ Sum of observations }}{\text{ Number of observations }}$ $=\frac{32+41+27+54+36+25+28+57+40+38}{10}$ The three central values of the data are: mean, median and mode. $=\frac{378}{10}$ years $=37.8$ years

Q. The mean age of 5 children of a family is 12 years. If four of them are respectively $6,11,13$ and 16 years, find the age of the fifth child. Solution: Let the age of the fifth child be x years. Then, mean age $=\frac{\text{ Sum of ages }}{\text{ Number of children }}$ $=\frac{6+11+13+16+\mathrm{x}}{5}$ $\Rightarrow 12=\frac{46+x}{5}$ $\Rightarrow \mathrm{x}+46=60$ $\Rightarrow x=60-46=14$ years.

Q. The mean of 5 observations is 15 . If mean of the first three observations is 14 and that of the last three is 17 , find the third observation. Solution: The mean of 5 observations $=15$ Total sum of 5 observations $=5\times 15=75$ Mean of first 3 observations $=14$ Total sum of first 3 observations $=3\times 14=42$ The mean of last 3 observations $=17$ Total sum of last 3 observations $=3\times 17=51$ $\therefore$ Third observation $=42+51-75=93-75=18$

Q. The runs scored by 11 members of a cricket team are $25,39,53,18,65,72,0,46,31,08,34$. Find the median score. Solution: Arranging the number of runs in ascending order, we have : $0,08,18,25,31,34,39,46,53,65,72$. Here $\mathrm{n}=11$, which is odd. $\therefore$ median score $=$ value of $\frac{1}{2}(11+1{)}^{\text{th }}$ term $=$ value of $6^{\text{th }}$ term $=34$. Hence, the median score is 34 .

Q. The weights of 10 students (in kg) are 40, 52, 34, 47, 31, 35, 48, 41, 44, 38. Find the median weight. Solution: Arranging the weights in ascending order, we have : 31, 34, 35, 38, 40, 41, 44, 47, 48, 52. Here, $\mathrm{n}=10$, which is even. $\therefore$ median weight $=\frac{1}{2}\times \{{\left(\frac{10}{2}\right)}^{\text{th }}$ term $+{\left(\frac{10}{2}+1\right)}^{\text{th }}$ term $\}$ $=\frac{1}{2}\{5^{\text{th }}$ term $+6^{\text{th }}$ term $\}$ $=\frac{1}{2}\{40+41\}\mathrm{kg}=\frac{81}{2}\mathrm{kg}=40.5\mathrm{kg}$ Hence, median weight $=40.5\mathrm{kg}$

Q. Find the mode of the following years of experience of teachers in a school: 10, 12, 5, 4, 7, 6, 7, 4, 2, 7, 1, 2, 3, 10, 1, 7, 5, 4. Solution: By inspection, the largest frequency is 4 . Therefore, 7 years of experience is the mode. More teachers have 7 years of experience than any other number of years.

Note that if the number of teachers with 7 years is only 3, and with 3 years is 2, then the second table would represent the situation.

The mode would now be 4 years and 7 years. When two scores appear with the same highest frequency we call such a frequency distribution bimodal.

Q. The number of cycles produced in a factory during five consecutive weeks is given below:

Draw a bar graph representing the above information. Solution: We can draw the bar graph by following these steps : Step 1. On a graph paper, draw a horizontal line OX and a vertical line $OY$, representing the $x$-axis and the $y$-axis respectively. Step 2. Along OX, mark the weeks at points taken at equal gaps. Step 3. Choose the scale : 1 small division $\equiv 200$ cycles. Step 4. The heights of the bars are : Production in the $1^{\text{st }}$ week $=\left(\frac{1}{200}\times 800\right)=4$ small divisions. Production in the $2^{\text{nd }}$ week $=\left(\frac{1}{200}\times 1300\right)=6.5$ small divisions. Production in the $3^{\text{rd }}$ week $=\left(\frac{1}{200}\times 1060\right)=5.3$ small divisions. Production in the $4^{\text{th }}$ week $=\left(\frac{1}{200}\times 920\right)=4.6$ small divisions. Production in the $5^{\text{th }}$ week $=\left(\frac{1}{200}\times 1440\right)=7.2$ small divisions. Step 5. Draw bars of equal width and of heights calculated in Step 4 at the points marked in Step 2. The completed bar graph is shown below.

Q. The graph shows the marks scored by Rahul in the four terminal exams. (i) In which term did Rahul score the most and how much ? (ii) What is his lowest score and in which term? (iii) What is the scale used to draw the graph ?

Q. The graph given below compares the heights and weights of various animals. The scale that is taken here is $1$ unit $=100\mathbf{cm}(\mathbf{or}\mathbf{kg})$

Q. A coin is tossed 100 times and head is obtained 59 times. On tossing a coin at random, find the probability of getting (i) a head (ii) a tail Solution: Total number of trials $=100$. Number of heads $=59$. Number of tails $=(100-59)=41$. (i) $\mathrm{P}($ getting a head $)=\frac{\text{ number of heads }}{\text{ total number of trials }}=\frac{59}{100}$ (ii) $\mathrm{P}($ getting a tail $)=\frac{\text{ number of tails }}{\text{ total number of trials }}=\frac{41}{100}$

Q. A dice is tossed 80 times and the number 3 is obtained 14 times. Now, a dice is tossed at random. Find the probability of getting the number 3. Solution: Total number of trials $=80$. Number of times 3 is obtained $=14$. $\therefore \mathrm{P}($ getting 3$)=\frac{\text{ number of times }3\text{ is obtained }}{\text{ total number of trials }}=\frac{14}{80}=\frac{7}{40}$.

Q. A dice is thrown 200 times and the outcomes are noted as shown below:

Q. When a dice is thrown at random, find the probability of getting a (i) 5 (ii) 3 (iii) 4 (iv) 6 Solution: Total number of trials $=200$. (i) $\mathrm{P}($ getting 5$)=\frac{29}{200}$. (ii) $\mathrm{P}($ getting 3$)=\frac{42}{200}=\frac{21}{100}$. (iii) $\mathrm{P}($ getting 4$)=\frac{38}{200}=\frac{19}{100}$. (iv) $\mathrm{P}($ getting 6$)=\frac{40}{200}=\frac{1}{5}$. (i)

(ii)

Q. The probability of selecting a queen from a standard pack of cards is $\frac{1}{13}$. Find the probability of not selecting a queen? Solution: $P($ event not happening $)=1-P($ event happening $)$ $P($ not selecting a queen $)=1-\frac{1}{13}=\frac{12}{13}$.

12.0Memory Map