Ex.14.1 Q4 Statistics Solution - NCERT Maths Class 10

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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 \(75.9\)

  
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