# Statistics - Class 9 - Mathematics

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## Extra Questions - Statistics | NCERT | Mathematics | Class 9

Variable | 1 | 2 | x | 4 | 5 |

Frequency | 2 | 3 | 4 | 5 | 6 |

The mean of the above frequency distribution is 3.5. Then the value of $x$ is:

A) 2

B) 3

C) 4

D) 5

The correct answer is option **B) 3**.

The problem states that the **mean** of the frequency distribution is $3.5$. We calculate the mean by using the formula:

$$ \text{Mean} = \frac{\sum_{i} (f_i x_i)}{\sum_{i} f_i} $$

Where $x_i$ represents the variables and $f_i$ their corresponding frequencies. Substituting the given values into the formula, we have:

$$ 3.5 = \frac{1 \cdot 2 + 2 \cdot 3 + x \cdot 4 + 4 \cdot 5 + 5 \cdot 6}{2+3+4+5+6} $$

Calculating the values in the numerator:

$1 \cdot 2 = 2$

$2 \cdot 3 = 6$

$x \cdot 4 = 4x$

$4 \cdot 5 = 20$

$5 \cdot 6 = 30$

Thus,

$$ 3.5 = \frac{2 + 6 + 4x + 20 + 30}{20} $$

which simplifies to

$$ 3.5 = \frac{58 + 4x}{20} $$

To find $x$, solve the equation:

$$ 70 = 58 + 4x $$

Resulting in

$$ 4x = 12 $$

Hence,

$$ x = 3 $$

This means the value of **x** in the distribution must be **3** to achieve the given mean of $3.5$.

Concentration of $\mathrm{SO}_2$ (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 |

The above frequency distribution table represents the concentration of Sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of Sulphur dioxide in the interval 0.12-0.16 on any of these days.

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The number of cars manufactured each day of a week are 2, 4, 6, 8, 10, 5, 7. Find the standard deviation of the number of cars manufactured using the Direct method.

A) 2.83

B) $\quad$ 3.45

C) 6.35

D) 8.23