Chapter 13: Statistics

Ans: The number of houses denoted by \[{{x}_{i}}\].

The mean can be found as given below:

\[\overline{X}=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:\[xi=\text{ }\frac{\text{Upper class limit+Lower class limit}}{2}\] 

\[{{x}_{i}}\] and \[{{f}_{i}}{{x}_{i}}\] can be calculated as follows:

From the table, it can be observed that

$\sum{{{f}_{i}}=20}$ 

$\sum{{{f}_{i}}{{x}_{i}}}=162$ 

Substituting the value of  \[{{f}_{i}}{{x}_{i}}\]and ${{f}_{i}}$ in the formula of mean we get:

Mean number of plants per house \[\left( \overline{X} \right)\]:

$   \overline{X}=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}} $

$  \overline{X}=\frac{162}{20}=8.1$        

Therefore, the mean number of plants per house is \[8.1\].

In this case, we will use the direct method because the value of \[{{x}_{i}}\] and ${{f}_{i}}$ .

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Take the assured mean \[(a)\] of the given data 

$a=150$

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

$ h=120-100 $

$ h=20 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be evaluated as follows:

From the table, it can be observed that

\[\sum{{{f}_{i}}=50}\] 

and

$\sum{{{f}_{i}}{{u}_{i}}}=-12$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h $

$\Rightarrow  \overline{X}=150+\left( \frac{-12}{50} \right)20 $

$ \Rightarrow  \overline{X}=150-\frac{24}{5} $

$ \Rightarrow  \overline{X}=150-4.8 $

$\overline{X}=145.2$

Hence, the mean daily wage of the workers of the factory is Rs \[145.20\].

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is $18$

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\text{ }\frac{\text{Upper class limit+Lower class limit}}{2}\] 

It is given that, mean pocket allowance, \[\overline{X}=\text{ }Rs\text{ }18\]

Class size (\[h\]) of this data is:

$h=13-11 $

$ h=2 $ 

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\]can be evaluated as follows:

From the table, it can be observed that

\[\sum{{{f}_{i}}=44+f}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=2f-40$ 

Substituting the value of \[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}} \right)h$ 

$\Rightarrow  18=18+\left( \frac{2f-40}{44+f} \right)2$ 

$\Rightarrow  0=\left( \frac{2f-40}{44+f} \right)$

$\Rightarrow  2f-40=0$ 

$\Rightarrow  f=20$ 

Hence, the value of frequency \[{{f}_{i}}\] is \[20\].

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is $75.5$

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

$ h=68-65 $

$h=3 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\]can be evaluated as follows:

From the table, it can be observed that

\[\sum{{{f}_{i}}=30}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=4$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h $

$ \Rightarrow  \overline{X}=75.5+\left( \frac{4}{30} \right)3 $

$ \Rightarrow  \overline{X}=75.5+0.4 $

$ \Rightarrow  \overline{X}=75.9 $

Therefore, mean heart beats per minute for these women are \[75.9\] beats per minute.

Ans:

It can be noticed that class intervals are not continuous in the given data. There is a gap of \[1\]  between two class intervals. Therefore, we have to subtract $\frac{1}{2}$ to lower class and have to add $\frac{1}{2}$ to upper to make the class intervals continuous.

Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is $57$.

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

$ h=52.5-49.5 $

$ h=3 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be calculated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=400}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=25$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

$ \Rightarrow  \overline{X}=57+\left( \frac{25}{400} \right)3 $

$ \Rightarrow  \overline{X}=57+\frac{3}{16} $

$ \Rightarrow \overline{X}=57.1875 $

Hence, mean number of mangoes kept in a packing box is $57.1875$.

In the above case, we used step deviation method as the values of \[{{f}_{i}},\text{ }{{d}_{i}}\] are large and the class interval is not continuous.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is \[225\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

$ h=150-100 $

$ h=50 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be calculated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=25}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-7$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$  \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h $

$\Rightarrow   \overline{X}=225+\left( \frac{-7}{25} \right)\times 50 $

 $\Rightarrow  \overline{X}=221 $

Hence, mean daily expenditure on food is Rs\[211\].

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is \[0.14\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

$ h=0.04-0.00 $

$ h=0.04 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be calculated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=30}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-31$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h $ 

$ \overline{X}=0.14+\left( \frac{-31}{30} \right)\times (0.04) $

$ \overline{X}=0.14-0.04133 $

$ \overline{X}=0.09867 $

$\overline{X}=0.099$ ppm

Hence, the mean concentration of \[\text{S}{{\text{O}}_{\text{2}}}\] in the air is $0.099$ppm.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)$

Suppose the assured mean $\left( a \right)$ of the data is \[17\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be calculated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=40}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-181$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right) $

$ \Rightarrow  \overline{X}=17+\left( \frac{-181}{40} \right) $

$ \Rightarrow  \overline{X}=17-4.525 $

$ \Rightarrow  \overline{X}=12.475 $

Hence, the mean number of days is $12.48$ days for which a student was absent.

Ans: Let the class size of the data be $h$.

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)$

Suppose the assured mean $\left( a \right)$ of the data is \[70\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

\[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\]

Class size (\[h\]) of this data is:

$ h=55-45 $

$ h=10 $

\[{{d}_{i}},\]\[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] can be calculated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=35}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-2$ 

Substituting the value of \[{{u}_{i}},\] and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean we get:

The required mean:

$ \overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h $

$ \Rightarrow  \overline{X}=70+\left( \frac{-2}{35} \right)\times 10 $

 $ \Rightarrow  \overline{X}=70-\frac{20}{35} $ 

 $\Rightarrow  \overline{X}=69.43 $

Therefore, the mean literacy rate of cities is $69.43%$.