# Exercise 14.2 Statistics NCERT Solutions Class 9

Exercise 14.2

## Chapter 14 Ex.14.2 Question 1

The blood groups of \(30\) students of Class VIII are recorded as follows:

\(A,\, B,\, O,\, O,\, AB,\, O,\, A,\, O,\, B,\, A,\, \\O,\, B,\, A,\, O,\, O, A,\, AB,\, O,\, A,\, A,\,\\ O,\, O,\, AB,\, B,\, A,\, O,\, B,\, A,\, B,\, O.\)

Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

**Solution**

**Video Solution**

**What is known?**

The blood group of \(30\) students of class** **VIII.

**What is unknown?**

The most common blood group and the rarest blood group.

**Reasoning:**

By drawing frequency distribution table, we can check which group is occurring most and which group is occurring least time.

**Steps:**

Let us sort the data using a table so as to make the data more easily understandable and its important features visible.

This table is known as frequency distribution table.

‘Frequency’ of a particular data refer to the number of times the data value occurs.

In our case frequency refer to the number of students having the same blood group.

\(\therefore\) The blood group of \(30\) students can be shown as follows:

Blood Group |
No of Students(frequency) |

\(A\) | \(9\) |

\(B\) | \(6\) |

\(O\) | \(12\) |

\(AB\) | \(3\) |

Total |
\(30\) |

So we can see easily now, from the table that:

- The most common (the most frequently occurring) blood group is ‘\(O\)’.
- The rarest blood group (least frequently occurring) is ‘\(AB\)’.

## Chapter 14 Ex.14.2 Question 2

The distance (in km) of \(40\) engineers from their residence to their place of work were found as follows:

\(5\,\,\,\, 3\,\,\,\, 10\,\,\,\, 20\,\,\,\, 25\,\,\,\, 11\,\,\,\, 13\,\,\,\, 7\,\,\,\, 12\,\,\,\, 31\\ 19\,\,\,\, 10\,\,\,\, 12\,\,\,\, 17\,\,\,\, 18\,\,\,\, 11\,\,\,\, 32\,\,\,\, 17\,\,\,\, 16\\\,\, 2 7\,\,\,\, 9\,\,\,\, 7\,\,\,\, 8\,\,\,\, 3\,\,\,\, 5\,\,\,\, 12\,\,\,\, 15\,\,\,\, 18\,\,\,\, 3\\ 12\,\,\,\, 14\,\,\,\, 2\,\,\,\, 9\,\,\,\, 6\,\,\,\, 15\,\,\,\, 15\,\,\,\, 7\,\,\,\, 6\,\,\,\, 12\)

Construct a grouped frequency distribution table with class size \(5\) for the data given above taking the first interval as \(0-5\) (\(5\) not included). What main features do you observe from this tabular representation?

**Solution**

**Video Solution**

**What is known?**

- The distance in \(\rm km\) of \(40\) engineers from their residence to work.
- Class size of \(5\) [ class size refers to the count of data that has to be accommodated in each grouping of data]
- Class interval as \(0-5\) (\(5\) not included).

**What is unknown?**

Main features of the tabular representation to be drawn from the given data.

**Reasoning:**

By drawing frequency distribution we can observe main features

**Steps:**

Since the given data is large, we need to group the data into classes of size each \(5\) and construct a table which will make the important features of the data visible.

This table is known as grouped frequency distribution table.

The class intervals will be \(0-5, 5-10, 10-15\) and so on.

The distance (in \(\rm km\)) of \(40\) engineers from their residence to their place work can be represented as follows (with the help of tally marks.)

DISTANCE (in \(\rm km\)) |
TALLY MARKS |
NO OF ENGINEERS (frequency) |

\(0-5 \) | \(5 \) | |

\(5-10 \) | \(11\) | |

\(10-15\) | \(11\) | |

\(15-20 \) | \(9 \) | |

\(20-25 \) | | | \(1 \) |

\(25-30 \) | | | \(1\) |

\(30-35\) | || | \(2\) |

Total |
\(40\) |

The following features can be observed from the table.

- \(5\) engineers have their houses below \(5\,\rm km\) distance from work place.
- A majority of engineers (\(36\)) have their houses below \(20\,\rm km\) distance from work place.
- Only a few engineers (\(4\)) have their houses at \(20\,\rm km\) and above distance from work place.

## Chapter 14 Ex.14.2 Question 3

The relative humidity (in \(\%\)) of a certain city for a month of \(30\) days was as follows:

\(98.1\,\,\, 98.6\,\,\, 99.2\,\,\, 90.3\,\,\, 86.5\,\,\, 95.3\,\,\, 92.9\,\,\,\\ 96.3\,\,\, 94.2\,\,\, 95.1\,\,\, 89.2\,\,\, 92.3\,\,\, 97.1\,\,\, 93.5\,\,\,\\ 92.7\,\,\, 95.1\,\,\,97.2\,\,\, 93.3\,\,\, 95.2\,\,\, 97.3\,\,\, 96.2\,\,\, \\92.1\,\,\, 84.9\,\,\, 90.2\,\,\, 95.7\,\,\, 98.3\,\,\, 97.3\,\,\, 96.1\,\,\,\\ 92.1\,\,\, 89\)

- Construct a grouped frequency distribution table with classes \(84 - 86, 86–88,\) etc.
- Which month or season do you think this data is about?
- What is the range of this data?

**Solution**

**Video Solution**

**What is known?**

- The relative humidity (in \(\%\)) of a certain city over a month of \(30\) days.
- Class size is \(2\), so the class intervals will be \(84-86, 86-88, 88-90\) and so on.

**What is unknown?**

- Constructing a grouped frequency distribution table.
- The month / season that the data talks about.
- Range of this data [which is the difference of the highest and the lowest values in the data.]

**Reasoning:**

By drawing frequency distribution table we can observe data month.

Range of data \(=\) maximum value \(–\) minimum value

**Steps:**

Construct a grouped frequency distribution table with class size of \(2\)

The relation humidity (in \(\%\)) of a certain city for a month can be represented as follows:

Relative Humidity (in \(\%\)) |
Number of Days (frequency) |

\(84-86\) | \(1\) |

\(86-88\) | \(1\) |

\(88-90\) | \(2\) |

\(90-92\) | \(2\) |

\(92-94\) | \(7\) |

\(94-96\) | \(6\) |

\(96-98\) | \(7\) |

\(98-100\) | \(4\) |

Total |
\(30\) |

The following features can be observed:

- The relative humidity was \(92\%\) and above, over a period of \(24 \)days \(\begin{align}(\frac{4}{5} \text{of a month})\end{align}\)
- Since the relative humidity \(\%\) is very high, it must be a data from a month of rainy season.

Range of data \(=\) Maximum value \(-\) Minimum value

\[\begin{align} &= 99.2 – 84.9\\ &= 14.3\end{align}\]

## Chapter 14 Ex.14.2 Question 4

The heights of \(50\) students, measured to the nearest centimeters, have been found to be as follows:

\(161\,\,\,\, 150\,\,\,\, 154\,\,\,\, 165\,\,\,\, 168\,\,\,\, 161\,\,\,\, 154\,\,\,\, 162\,\,\,\, \\150\,\,\,\, 151\,\,\,\,162\,\,\,\, 164\,\,\,\, 171\,\,\,\, 165\,\,\,\, 158\,\,\,\, 154\,\,\,\,\\ 156\,\,\,\, 172\,\,\,\, 160\,\,\,\, 170\,\,\,\, 153\,\,\,\, 159\,\,\,\, 161\,\,\,\, 170\,\,\,\,\\ 162\,\,\,\, 165\,\,\,\, 166\,\,\,\, 168\,\,\,\, 165\,\,\,\,164\,\,\,\, 154\,\,\,\, 152\,\,\,\,\\ 153\,\,\,\, 156\,\,\,\, 158\,\,\,\, 162\,\,\,\, 160\,\,\,\, 161\,\,\,\, 173\,\,\,\, 166 \,\,\,\,\\161\,\,\,\, 159\,\,\,\, 162\,\,\,\, 167\,\,\,\, 168\,\,\,\, 159\,\,\,\, 158\,\,\,\, 153\,\,\,\,\\ 154\,\,\,\, 159\)

- Represent the data given above by a grouped frequency distribution table,Taking the class intervals as \((160 – 165), (165 – 170),\) etc.
- What can you conclude about their heights from the table?

**Solution**

**Video Solution**

**What is known?**

- The heights of \(50\) students in \(\rm cm.\)
- Class size is \(5\) [from the given instruction of class intervals to be taken as \(160- 165, 165-170, 170-175\) …. and so on].

**What is unknown?**

- Constructing a grouped frequency distribution table.
- Making inferences and drawing conclusions from the data of “heights of students”.

**Reasoning:**

According to class interval we can check how many numbers lie in between , like this we can draw frequency distribution table and can be conclude about height from table.

**Steps:**

A grouped frequency distribution table with a class size of \(5,\) for the heights of \(50\) students can be constructed as follows:

Height (in cm) |
Number of Students (frequency) |

\(150-155\) | \(12\) |

\(155-160\) | \(9\) |

\(160-165\) | \(14\) |

\(165-170\) | \(10\) |

\(170-175\) | \(5\) |

Total |
\(50\) |

Since ‘\(150\)’ is the lowest value, we started the \(1\rm st\) class interval at \(150.\)

The following features / conclusions can be drawn from the table:

- A majority of \(70\%\) of the students (\(35\) students) are below \(165\rm cm\) height.
- Only a \(10\%\) of the students (\(5\) students) are at a height of \(170\rm cm\) or above.

## Chapter 14 Ex.14.2 Question 5

A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

\[\begin{align}0.03\;\;0.08\;\; 0.08\;\; 0.09\;\; 0.04\;\; 0.17 \\0.16\;\; 0.05\;\; 0.02\;\; 0.06\;\; 0.18\;\; 0.20 \\

0.11 \;\;0.08 \;\;0.12\;\; 0.13\;\; 0.22\;\; 0.07\;\; \\0.08\;\; 0.01\;\; 0.10\;\; 0.06\;\; 0.09\;\; 0.18 \\0.11\;\; 0.07\;\; 0.05\;\; 0.07\;\; 0.01 \;\;0.04\end{align}\]

- Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.
- For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

**Solution**

**Video Solution**

**What is known?**

- Concentration of sulphur dioxide in the air (in ppm) of a certain city observed over \(30\) days
- Class intervals of ‘\(0.00 – 0.04’, ‘0.04 - 0.08\)’, and so on. Hence class size is \(0.04.\)

**What is unknown? **

Constructing a grouped frequency table for the given data which will help us to find out for how many days was the concentration of sulphur dioxide more than \(0.11\) ppm.

**Reasoning:**

According to class interval we can check how many numbers lie in between, like this we can draw frequency distribution table and can be conclude about no. of days in which Sulphur dioxide more than \(0.11\) parts per million from table.

**Steps:**

A grouped frequency distribution table with a class size of \(‘0.04\)’ needs to be constructed for the given data.

CONCENTRATION OF SULPHUR DI OXIDE (in ppm) |
NUMBER OF DAYS (frequency) |

\(0.00 - 0.04\) | \(4\) |

\(0.04 - 0.08\) | \(9\) |

\(0.08 - 0.12\) | \(9\) |

\(0.12 - 0.16\) | \(2\) |

\(0.16 - 0.20\) | \(4\) |

\(0.20 - 0.24\) | \(2\) |

Total |
\(30\) |

From this table we can see that the number of days during which the concentration of Sulphur dioxide is more than \( 0.11 \)ppm, falls over three class intervals, \(0.12-0.16, 0.16-0.20\) and \(0.20-0.24.\)

So,\(2+4+2\text{ }\Rightarrow \text{ }8\)

\(8\) days had a concentration of Sulphur dioxide more than \(0.11\) ppm.

## Chapter 14 Ex.14.2 Question 6

Three coins were tossed \(30\) times simultaneously. Each time the number of Heads occurring was noted down as follows:

\[\begin{align}0\,\, 1\,\,\, 2\,\,\, 2\,\,\, 1\,\,\, 2\,\,\, 3\,\,\, 1\,\,\, 3 \,\,\,0\,\,\,\\ 1\,\,\, 3\,\,\, 1\,\,\, 1\,\,\, 2\,\,\, 2\,\,\, 0\,\,\, 1\,\,\, 2\,\,\, 1\,\,\\ 3\,\, 0\,\,\, 0\,\,\, 1\,\,\, 1\,\,\, 2\,\,\, 3\,\,\, 2\,\,\, 2\,\,\, 0\end{align}\]

Prepare a frequency distribution table for the data given above.

**Solution**

**Video Solution**

**What is known?**

- Number of times three coins were tossed simultaneously.
- Number of ‘heads’ that occurred during each toss.

**What is unknown?**

Constructing the required frequency distribution table.

**Reasoning:**

We can count in each case and can be written in front of them , like this we can draw frequency distribution table.

**Steps:**

By observing the given data, we can prepare the frequency distribution table as follows:

NUMBER OF HEADS |
TALLY MARKS |
FREQUENCY OF OCCURRENCE |

\(0 \) | \(6 \) | |

\(1 \) | \(10 \) | |

\(2 \) | \(9\) | |

\(3\) | \(5\) | |

TOTAL |
\(30\) |

## Chapter 14 Ex.14.2 Question 7

The value of \(π\) upto \(50\) decimal places is given below:

\(3.14159265358979323846264338327950288419716939937510\)

(i) Make a frequency distribution of the digits from \(0\) to \(9\) after the decimal point.

(ii) What are the most and the least frequently occurring digits?

**Solution**

**Video Solution**

**What is known?**

- The value of ‘\(π\)’ upto \(50\) decimel places.

**What is unknown?**

- Constructing a required frequency distribution table
- And finding the most and the least frequently occurring digits.

**Reasoning: **

We can count each digit and can be written in front of them , like this we can draw frequency distribution table and can be concluded about most or least frequently occurring digits.

**Steps:**

We can represent the data by constructing a simple, ungrouped frequency distribution table as follows.

DIGITS |
TALLY MARKS |
FREQUENCY OF OCCURRENCE |

\(0 \) | || | 2 |

\(1 \) | 5 | |

\(2 \) | 5 | |

3 | 8 | |

4 | |||| | 4 |

5 | 5 | |

6 | |||| | 4 |

7 | |||| | 4 |

8 | 5 | |

9 | 8 | |

TOTAL |
50 |

If can be easily observed from the table that

- The most frequently occurring digits are \(3\) and \(9\) with a max frequency of \(8\) and
- The least frequently occurring digit is \(‘0’\) with a lowest frequency of \(2\).

## Chapter 14 Ex.14.2 Question 8

Thirty children were asked about the number of hours they watched TV Programmes in the previous week. The results were found as follows:

\[\begin{align}1\,\,\,\, 6\,\,\,\, 2\,\,\,\, 3\,\,\,\, 5\,\,\,\, 12\,\,\,\, 5\,\,\,\, 8\,\,\,\, 4\,\,\,\, 8\,\,\,\, \\10\,\,\,\, 3\,\,\,\, 4\,\,\,\, 12\,\,\,\, 2\,\,\,\, 8\,\,\,\, 15\,\,\,\, 1\,\,\,\, 17\,\,\,\, 6\,\,\,\, \\3\,\,\,\, 2\,\,\,\, 8\,\,\,\, 5\,\,\,\, 9\,\,\,\, 6\,\,\,\, 8\,\,\,\, 7\,\,\,\, 14\,\,\,\, 12\,\,\,\,\end{align}\]

(i) Make a grouped frequency distribution table for this data, taking class width \(5\) and one of the class intervals as \(5 - 10.\)

(ii) How many children watched television for \(15\) or more hours a week?

**Solution**

**Video Solution**

**What is known?**

- Number of hours of watching TV programs by \(30\) children in a weak.
- Class width ‘\(5\)’
- A Class interval, such as \(5-10\)

**What is unknown?**

- A grouped frequency distribution table with a class width of ‘\(5\)’
- Number of children who watched television for \(15\) or more hours per weak

**Steps:**

A grouped frequency distribution table can be constructed as follows with the class intervals of \(0-5, 5-10, 10-15, 15-20\) and so on.

HOURS |
NUMBERS OF CHILDREN (frequency) |

\(0-5 \) | \(10\) |

\(5-10 \) | \(13\) |

\(10-15 \) | \(5 \) |

\(15-20\) | \(2\) |

TOTAL |
\(30\) |

We can observe from the table, that the number of children who watched television for \(15\) or more hours a week is \(2\) (which falls under the class interval ’\(15-20\)’)

## Chapter 14 Ex.14.2 Question 9

A company manufactures car batteries of a particular type. The lives (in years) of \(40\) such batteries were recorded as follows:

\(2.6\) | \(3.0\) | \(3.7\) | \(3.2\) | \(2.2\) | \(4.1\) | \(3.5\) | \(4.5\) |

\(3.5\) | \(2.3\) | \(3.2\) | \(3.4\) | \(3.8\) | \(3.2\) | \(4.6\) | \(3.7\) |

\(2.5\) | \(4.4\) | \(3.4\) | \(3.3\) | \(2.9\) | \(3.0\) | \(4.3\) | \(2.8\) |

\(3.5\) | \(3.2\) | \(3.9\) | \(3.2\) | \(3.2\) | \(3.1\) | \(3.7\) | \(3.4\) |

\(4.6\) | \(3.8\) | \(3.2\) | \(2.6\) | \(3.5\) | \(4.2\) | \(2.9\) | \(3.6\) |

Construct a grouped frequency distribution table for this data, using class Intervals of size \(0.5\) starting from the interval \(2 - 2.5.\)

**Solution**

**Video Solution**

**What is known?**

- The life span (in years) of \(40\) car batteries manufactured by a company.
- Class size is ‘\(0.5\)’
- A class interval such as \(2-2.5\)

**What is unknown?**

- Constructing a grouped frequency distribution table with a class size of \(0.5.\)

**Steps:**

The required grouped frequency distribution table can be constructed as follows, with class intervals of ‘\(2-2.5\)’, ‘\(2.5-3\)’, ‘\(3-3.5\)’, ‘\(3.5-4\)’, and so on.

LIFE SPAN OF BATTERIES (in years) |
NUMBER OF BATTERIES (frequency) |

\(2-2.5 \) | \(2\) |

\(2.5-3.0 \) | \(6 \) |

\(3.0-3.5\) | \(14 \) |

\(3.5-4.0 \) | \(11 \) |

\(4.0-4.5 \) | \(4 \) |

\(4.5-5.0\) | \(3\) |

TOTAL |
\(40\) |