# 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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