# Exercise 14.4 Statistics NCERT Solutions Class 9

Exercise 14.4

## Question 1

The following number of goals were scored by a team in a series of \(10\) matches:

\(2, 3, 4, 5, 0, 1, 3, 3, 4, 3\)

Find the mean, median and mode of these scores.

### Solution

**Video Solution**

**What is known?**

Number of goals was scored by a team in a series of \(10\) matches

**What is unknown?**

Mean, median and mode.

**Reasoning:**

The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations.

The median is that value of the given number of observations, which divides it into exactly two parts. So, when the data is arranged in ascending (or descending) order the median of ungrouped data can be calculated based on no. of observation are even or odd.

The mode is that value of the observation which occurs most frequently.

**Steps:**

The number of goals scored by the team is

\(2, 3, 4, 5, 0, 1, 3, 3, 4, 3\)

\(\overline{\rm{Mean}}\,{\rm{of }}\,{\rm{data}}\)

\[\begin{align}&=\frac{\text { Sum of all the observations }}{\text { Total number of observations }}\end{align}\]

\(\overline{\rm{Mean}}\,{\rm{score}}\)

\[\begin{align}&=\frac{2\!+\!3\!+\!4\!+\!5\!+\!0\!+\!1\!+\!3\!+\!3\!+\!4\!+\!3}{10}\\ \\ &=\frac{28}{10} \\ &=2.8 \\&= \,2.8\,\,{\rm{goals}}\end{align}\]

Arranging them in ascending order we get

\(0, 1, 2, 3, 3, 3, 3, 4, 4, 5\)

The number of observations is \(10\), which is an even number. Therefore, median score will be the mean of\(\frac{{10}}{2}\) _{i.e.,}\( 5\)^{ th} \(\frac{{10}}{2}\) and \(+ 1\) i.e., \(6\) th observation while arranged in ascending or descending order.

Median score

\[\begin{align}&= \frac{{{5^{th}}\,{\text{observation}}\,{\rm{ + }}\,{6^{{\rm{th}}}}{\text{observation}}\,}}{2}\\&= \frac{{3 + 3}}{2}\\ &= \frac{6}{2}\\ &= 3\end{align}\]

Mode of data is the observation with the maximum frequency in data.

Therefore, the mode score of data is \(3\) as it has the maximum frequency as \(4\) in the data.

## Question 2

In a mathematics test given to \(15\) students, the following marks (out of \(100\)) are recorded:

\(41, 39, 48, 52, 46, 62, 54, \\ 40, 96, 52, 98, 40, 42, 52, 60.\)

Find the mean, median and mode of this data.

### Solution

**Video Solution**

**What is known?**

Marks obtained by \(15\) students in mathematics test

**What is unknown?**

Mean, median and mode.

**Reasoning:**

The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations.

The median is that value of the given number of observations, which divides it into exactly two parts. So, when the data is arranged in ascending (or descending) order the median of ungrouped data can be calculated based on no. of observation are even or odd.

The mode is that value of the observation which occurs most frequently.

**Steps:**

The marks obtained by \(15\) students in mathematics test:

\(41, 39, 48, 52, 46, 62, 54, \\40, 96, 52, 98, 40, 42, 52, 60.\)

Mean of data

\(\begin{align} &= \frac{ \text{sum of all observation}}{ \text{Total number of observation}} \\\\&= \frac{\begin{Bmatrix} 41 + 39 + 48 + 52 +\\ 46 + 62 + 54 + 40 +\\ 96 + 52 + 98 + 40 \\ + 42 + 52 + 60 \end{Bmatrix} }{15}\\&= \frac{822}{15}\\\\ &= 54.8\end{align}\)

Arranging them in ascending order we get:

\(39,40, 40, 41, 42, 46, 48, 52,\\ 52, 52, 54, 60, 62, 96, 98. \)

As the number of observations is \(15\) which is odd, therefore, the median of data will be \(\begin{align} \frac{{15 + 1}}{2}\end{align} = \) \(8\) ^{th }observation whether the data is arranged in an ascending or descending order.

Therefore, median score of data \(= 52\)

Mode of data is the observation with the maximum frequency in data. Therefore, mode of this data is \(52\) having the highest frequency in data as \(3\).

## Question 3

The following observations have been arranged in ascending order. If the median of the data is \(63,\) find the value of \(x.\)

\(29, 32, 48, 50, x, x + 2, 72, 78, 84, 95\)

### Solution

**Video Solution**

**What is known?**

Ungrouped data and median of data is \(63.\)

**What is unknown?**

Value of \(x.\)

**Reasoning:**

The median is that value of the given number of observations, which divides it into exactly two parts. So, when the data is arranged in ascending (or descending) order the median of ungrouped data can be calculated based on no. of observation are even or odd.

**Steps:**

It can be observed that the total number of observations in the given data is \(10\) (even number).

Therefore, the median of this data will be the mean of \(\begin{align} \frac{{10}}{2}\end{align} \) i.e.,_{ \(5\)}^{th }and \(\begin{align} \frac{{10}}{2}\end{align} \) \(+ 1\) i.e., \(6\)^{th }observation.

Therefore, median of the data

\(\begin{align} &= \frac{ \begin{Bmatrix} 5^{\rm th} \text{observation } + \\ 6^{\rm th} \text{ observation} \end{Bmatrix} }{2}\\& \Rightarrow \,\,63 = \,\frac{{x + x + 2}}{2}\\&\Rightarrow \,\,63 = \,\frac{{2x + 2}}{2}\,\\&\Rightarrow \,\,63\, = \,x + 1\\& \Rightarrow \,\,x\, = \,62\end{align}\)

## Question 4

Find the mode of

\(14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.\)

### Solution

**Video Solution**

**What is known?**

Ungrouped data.

**What is unknown?**

Mode of the given data.

**Reasoning:**

The mode is that value of the observation which occurs most frequently.

**Steps:**

Arranging the given data in the ascending order:

\(14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28 \)

It can be observed that ‘\(14\)’ repeats \(4\) times.

Therefore, mode of the given data is \(14.\)

## Question 5

Find the mean salary of \(60\) workers of a factory from the following table:

Salary (in Rs.) |
Number of Workers |

\(3000\) | \(16\) |

\(4000\) | \(12\) |

\(5000\) | \(10\) |

\(6000\) | \(8\) |

\(7000\) | \(6\) |

\(8000\) | \(4\) |

\(9000\) | \(3\) |

\(10000\) | \(1\) |

Total |
60 |

### Solution

**Video Solution**

**What is known?**

Salary of \(60\) workers in tabular form

**What is unknown?**

Mean of the given data

**Steps:**

We know that

Mean \(=\) \(\begin{align} \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\end{align} \)

The value of \(\sum {{f_i}{x_i}} \) and \(\sum {{f_i}} \) f can be calculated as follows.

Salary \((x_{i})\) |
Number of Workers \((f_{i})\) |
\(f_{i} x_{i}\) |

\(3000\) | \(16\) | \(\begin{align}16 \times 3000\\ =48000 \end{align} \) |

\(4000\) | \(12\) | \(\begin{align} 12 \times 4000 \\ =48000 \end{align}\) |

\(5000\) | \(10\) | \(\begin{align} 10 \times 5000 \\ =50000 \end{align} \) |

\(6000\) | \(8\) | \( \begin{align} 8 \times 6000 \\ =48000 \end{align} \) |

\(7000\) | \(6\) | \( \begin{align} 6 \times 7000 \\ =42000 \end{align} \) |

\(8000\) | \(4\) | \( \begin{align} 4 \times 8000 \\ =32000 \end{align}\) |

\(9000\) | \(3\) | \( \begin{align} 3 \times 9000 \\ =27000 \end{align} \) |

\(10000\) | \(1\) | \( \begin{align} 1 \times 10000 \\ =10000 \end{align} \) |

Total |
\(\sum f_{i}=60\) |
\(\sum f_{i} x_{i}=305000\) |

Mean salary \(\begin{align} = \,\frac{{305000}}{{60}}\end{align} \)

Therefore, mean salary of \(60\) workers is \(\rm{Rs.}\, 5083.33. \)

## Question 6

Give one example of a situation in which

(i) the mean is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

### Solution

**Video Solution**

**Reasoning:**

Extreme values in the data affect the mean. This is one of the weaknesses of the mean. So, if the data has a few points which are very far from most of the other points, (like \(1,7,8,9,9\)) then the mean is not a good representative of this data. Since the median and mode are not affected by extreme values present in the data, they give a better estimate of the average in such a situation.

**Steps:**

When any data has a few observations such that these are very far from the other observations in it, it is better to calculate the median than the mean of the data as median gives a better estimate of average in this case.

(i) Consider the following example − the following data represents the heights of the members of a family.

\(\begin{align} &154.9 \, \rm cm, 162.8 \,cm, 170.6 \, cm,\\ & 158.8 \, \rm cm, 163.3 \, \rm cm, 166.8 \, cm, \\ & 160.2 \, \rm cm \end{align} \)

In this case, it can be observed that the observations in the given data are close to each other.

Therefore, mean will be calculated as an appropriate measure of central tendency.

(ii) The following data represents the marks obtained by \(12\) students in a test.

\(\begin{align} & 48,59,46,52,54,46, \\ & 97,42,49,58,60,99\end{align} \)

In this case, it can be observed that there are some observations which are very far from other observations. Therefore, here median will be calculated as an appropriate measure of central tendency.

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school