Chapter 13: Statistics

Ans: 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 \[30\].

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 evaluated as follows:

It can be observed that from the above table

\[\sum{{{f}_{i}}=80}\] 

$\sum{{{f}_{i}}{{d}_{i}}}=430$ 

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) $ 

$  \overline{X}=30+\left( \frac{430}{80} \right) $

$  \overline{X}=30+5.375 $

$  \overline{X}=35.38 $

Hence, the mean of this data is $35.38$. It demonstrates that the average age of a patient admitted to hospital was $35.38$years.

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

It can be noticed that the maximum class frequency is \[23\]  belonging to class interval\[35\text{ }-\text{ }45\] .

Modal class \[=35\text{ }-\text{ }45\]

The values of unknowns is given as below as per given data:

\[l=\text{ }35\] 

\[{{f}_{1}}=\text{ }23\] 

\[h=15-5=10\] 

\[{{f}_{0}}=\text{ }21\] 

\[{{f}_{2}}=\text{ }14\] 

Substituting these values in the formula of mode we get:

$   M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=35+\left( \frac{23-21}{2(23)-21-14} \right)\times 10 $

$  M=35+\left[ \frac{2}{46-35} \right]\times 10 $

$  M=35+\frac{20}{11} $

$   \text{M=35+1}\text{.81} $

$  \text{M=36}\text{.8} $

Hence, the Mode of the data is $36.8$. It demonstrates that the age of maximum number of patients admitted in hospital was $36.8$years.

Ans: Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

From the data given above, it can be noticed that the maximum class

frequency is\[61\] belongs to class interval \[60\text{ }-\text{ }80\].

Therefore, Modal class \[=60-80\] 

The values of unknowns are given as below as per given data:

\[l=\text{60}\] 

\[{{f}_{1}}=61\] 

\[h=20\] 

\[{{f}_{0}}=52\] 

\[{{f}_{2}}=38\] 

Substituting these values in the formula of mode we get:

$   M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=60+\left( \frac{61-52}{2(61)-52-38} \right)\times 20 $

$  M=60+\left[ \frac{9}{122-90} \right]\times 20 $

$  M=360+\left( \frac{9\times 20}{32} \right) $

 $  M=60+\frac{90}{16}=60+5.625 $

$  M=65.625 $

Hence, the modal lifetime of electrical components is \[65.625\] hours.

Ans: For mode

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

It can be observed from the given data that the maximum class frequency is\[40\] ,

Belongs to \[1500\text{ }-\text{ }2000\] intervals.

Therefore, modal class \[=\text{ }1500\text{ }-\text{ }2000\] 

The values of unknowns are given as below as per given data:

\[l=1500\] 

\[{{f}_{1}}=40\] 

\[{{f}_{0}}=24\] 

\[{{f}_{2}}=33\] 

\[h=500\] 

Substituting these values in the formula of mode we get:

$   M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=1500+\left( \frac{40-24}{2(40)-24-33} \right)\times 500 $

$  M=1500+\left[ \frac{16}{80-57} \right]\times 500 $

$  M=1500+\frac{8000}{23} $

$M=1500+347.826$

$ M=1847.826 $

$  M\approx 1847.83 $

Therefore, modal monthly expenditure was Rs \[1847.83\] .

For mean,

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 \[2750\].

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

\[\text{Class mark}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

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

$   h=1500-1000 $

$ h=500 $

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

It can be observed from the above table

\[\sum{{{f}_{i}}=200}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-35$ 

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

The required mean:

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

$\overline{X}=2750+\left( \frac{-35}{200} \right)\times (500)$ 

$\overline{X}=2750-87.5$

$\overline{X}=2662.5$

Hence, the mean monthly expenditure is Rs$2662.5$ .

Ans: For mode

Mode can be calculated as,

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

It can be observed from the given data that the maximum class frequency is \[10\] which belongs to class interval\[30\text{ }-\text{ }35\] . 

Therefore, modal class \[=\text{ }30\text{ }-\text{ }35\] 

\[h=5\] 

\[l=30\] 

\[{{f}_{1}}=10\] 

\[{{f}_{0}}=9\] 

\[{{f}_{2}}=3\] 

Substituting these values in the formula of mode we get:

$   M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=30+\left( \frac{10-9}{2(10)-9-3} \right)\times 5 $

$  M=30+\left( \frac{1}{20-12} \right)\times 5 $

$  M=30.6 $

Hence, the mode of the given data is $30.6$. It demonstrates that most of the states/U.T have a teacher-student ratio of $30.6$ .

For mean,

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 \[32.5\].

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

\[\text{Class mark}=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

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

  $ h=20-15 $

$ h=5 $ 

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

It can be observed from the above table

\[\sum{{{f}_{i}}=35}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=-23$ 

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

The required mean

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

$  \overline{X}=32.5+\left( \frac{-23}{35} \right)\times 5 $

$  \overline{X}=32.5-\frac{23}{7} $

$  \overline{X}=29.22 $

Hence, the mean of the data is $29.2$ .

It demonstrates that the average teacher−student ratio was $29.2$.

Ans: Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

From the given data, it can be observed that the maximum class frequency is $18$ 

Belongs to class interval \[4000-5000\].

Therefore, modal class = \[4000-5000\]

\[l=4000\] 

\[{{f}_{1}}=18\] 

\[{{f}_{0}}=4\] 

\[{{f}_{2}}=9\] 

\[h=1000\] 

Substituting these values in the formula of mode we get:

$   M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=4000+\left( \frac{18-4}{2(18)-4-9} \right)\times 1000 $

$  M=4000+\left( \frac{14000}{23} \right) $

$  M=4608.695 $

Hence, the mode of the given data is \[4608.7\] runs.

Ans: Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

From the given data, it can be observed that the maximum class frequency is $20$,

Belonging to \[40-50\] class intervals.

Therefore, modal class = \[40-50\]

\[l=40\] 

\[{{f}_{1}}=20\] 

\[{{f}_{0}}=12\] 

\[{{f}_{2}}=11\] 

\[h=10\] 

Substituting these values in the formula of mode we get:

$  M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h $

$  M=40+\left( \frac{20-12}{2(20)-12-11} \right)\times 10 $

$  M=40+\left( \frac{80}{40-23} \right) $

$  M=44.7 $

Hence, the mode of this data is $44.7$ cars.