# Ex.14.1 Q4 Statistics Solution - NCERT Maths Class 10

## Question

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute |
\(65 – 68\) | \(68 – 71\) | \(71 – 74\) | \(74 – 77\) | \(77 – 80\) | \(80 – 83\) | \(83 – 86\) |

Number of women |
\(2\) | \(4\) | \(3\) | \(8\) | \(7\) | \(4\) | \(2\) |

## Text Solution

**What is known?**

The heart beats per minute of \(30 \) women.

**What is unknown?**

The mean heart beats per minute for these women.

**Reasoning:**

We will use Step-deviation Method to solve this question because the data given is large and will be convenient to apply to all the \(d_i\) have a common factor.

Sometimes when the numerical values of *\(x_i\)* and \(f_i \) are large, finding the product of *\(x_i\)* and \(f_i \)becomes tedious. We can do nothing with the \(f_i,\) but we can change each *\(x_i\)* to a smaller number so that our calculations become easy. Now we have to subtract a fixed number from each of these \(x_i.\)

The first step is to choose one among the \(x_i \)as the assumed mean and denote it by \(‘a’\). Also, to further reduce our calculation work, we may take \(‘a’\) to be that *\(x_i\)* which lies in the centre of \(x_1, x_2, . . ., x_n.\) So, we can choose \(a.\)

The next step is to find the difference \(‘d_i’\) between *a* and each of the \(x_i,\) that is, the deviation of \(‘a’\) from each of the \(x_i.\) i.e.,\(d_i = x_i - a\)

The third step is to find \(‘u_i’\) by dividing \(d_i\) and class size ** \(h\)** for each of the \(x_i.\) i.e.,\(u_i = \frac{{d_i}}{h}\)

The next step is to find the product of \(u_i\) with the corresponding \(f_i,\) and take the sum of all the \(f_iu_i\)

The step-deviation method will be convenient to apply if all the \(d_i \)have a common factor.

Now put the values in the below formula

Mean, \(\overline x = a + (\frac{{\Sigma f_i u_i}}{\Sigma f_i}) \times h\)

**Steps:**

We know that,

Class mark, \( x_i = \frac{{\text{Upper class limit + Lower class limit}}}{2}\)

Class size, \(h=3\)

Taking assumed mean,\( a= 75.5\)

Number of heart beats per minute |
No of women \( (f_i)\) |
\( X_i\) | \( d_i = x_i -a \) | \(\begin{align} {u_i}=\frac{ {x_i}-{a}}{h}\end{align}\) | \( f_iu_i \) |

\(65-68\) | \(2\) | \(66.5\) | \(-9\) | \(-3\) | \(-6\) |

\(68-71\) | \(4\) | \(69.5\) | \(-6\) | \(-2\) | \(-8\) |

\(71 -74\) | \(3\) | \(72.5\) | \(-3\) | \(-1\) | \(-3\) |

\(74-77\) | \(8\) | \(75.5(a)\) | \(0\) | \(0\) | \(0\) |

\(77 – 80\) | \(7\) | \(78.5\) | \(3\) | \(1\) | \(7\) |

\(80 – 83\) | \(4\) | \(81.5\) | \(6\) | \(2\) | \(8\) |

\(83 – 86\) | \(2\) | \(84.5\) | \(9\) | \(3\) | \(6\) |

\(\Sigma f_i=30 \) | \(\Sigma f_iu_i=4\) |

From the table, we obtain

\(\Sigma f_i=30\)

\(\Sigma f_i u_i=4\)

\[\begin{align} \operatorname{Mean}\,\,(\overline{{x}}) &={a}+\left(\frac{\Sigma {f}_{{i}} {u}_{{i}}}{\Sigma {f}_{{i}}}\right) {h} \\ \overline{{x}} &=75.5+\left(\frac{4}{30}\right) 3 \\ \overline{{x}} &=75.5-\frac{{2}}{5} \\\overline{{x}} &=75.5-0.4 \\ \overline {{x}}&=75.9 \end{align}\]

Hence, the mean heartbeat per minute for these women is \(75.9\)