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Ex.14.1 Q4 Statistics Solution - NCERT Maths Class 10

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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 \(2\) \(4\) \(3\) \(8\) \(7\) \(4\) \(2\)

 Video Solution
Ex 14.1 | Question 4

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.


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


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 \(75.9\)

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