NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Ans : To find the class mark \(  \left(x_{i}\right) \)  for each interval, the following relation is used.
Class mark \(  \left(x_{i}\right) \) \( \frac{\text { Upper class limit + Lower class limit }}{2}  \)
\(  x_{i} \text { and } f_{X_{i}} \) can be calculated as follows 

From the table it can be observed that 
\(  \sum f_{i}=20 \)
\(  \sum f_{r} x_{i}=162 \)
Mean \(  \overline{x}=\frac{\sum f_{f_{i}}}{\sum f_{i}} \)
\( =\frac{162}{20}=8.1  \)
Therefore mean number of plants per house is 8.1 Here, direct method has been used as the values of class marks \( \left(x_{i}\right)  \) and \( f_{i}  \) are small 

Ans : To find the class mark for each interval, the following relation is used,
\( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}  \)
Class size (h ) of this data =20
Taking 150 as assured mean (a) d, u and f  can be calculated as follows.

From the table it can be observed that 
\( \sum f_{i}=50  \)
\(  \sum f_{i} u_{i}=-12 \)
Mean \( \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h  \)
\(  =150+\left(\frac{-12}{50}\right) 20 \)
\(  =150-\frac{24}{5} \)
\(=150-4.8  \)
\(  =145.2 \)
Therefore the mean daily wage of the workers of the factory is Rs 145.20 

Ans : To find the class mark \(\left(x_{i}\right)   \) for each interval, the following relation is used.
\( x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}  \)
Given that mean pocket allowance \(  \overline{x}=\operatorname{Rs} 18 \)
Taking 18 as assured mean \( (a), d_{i} \text { and } f d_{i}  \) are calculated as follows. 

From the table, we obtain 
\(  \sum f_{i}=44+f \)
\(  \sum f_{i} u_{l}=2 f-40 \)
\(\overline{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{l}}   \)
\(  18=18+\left(\frac{2 f-40}{44+f}\right) \)
\( 0=\left(\frac{2 f-40}{44+f}\right) \)
\(  2 f-40=0 \)
\(  2 f=40 \)
\( f=20  \)
Hence the missing frequency f is 20 

Ans : To find the class mark of each interval (x,), the following relation is used.
\(  x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2} \)
Class size h of this data =3
Taking 75.5 as assumed mean \(  (a), d i, u_{i t}, f_{i} u_{i} \) are calculated as follows.

From the table  we obtain 
\(  \sum f i=30 \)
\(\sum f_{\mu,}=4  \)
Mean \( \overline{x}=a+\left(\frac{\sum f u_{i}}{\sum f_{i}}\right) \times h  \)
\(=75.5+\left(\frac{4}{30}\right) \times 3   \)
\( =75.5+0.4=75.9  \)
Therefore ,mean hear beats per minute for these women are 75.9 beats per minute. 

Ans : 
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit and 1/2 has to be subtracted from the lower class limit of each interval. 
Class mark (x,) can be obtained by using the following elation. 
\(  x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2} \)
Class size (h) of this data = 3 
Taking 57 as assumed mean , \((a), d_{i}, u_{i t} f_{\mu_{i}}   \) are calculated as follows. 

It can be observed that 
\( \sum f_{i}=400  \)
\(\sum f_{i} u_{i}=25  \)
Mean \(  \overline{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \)
\(=57+\left(\frac{25}{400}\right) \times 3   \)
\(  =57+\frac{3}{16}=57+0.1875 \)
\( =57.1875  \)
\(  =57.19 \)
Mean number of mangoes kept in a packing box is 57.19 Step deviation method is used here as the values of \(  f_{i,} d_{i} \) are big and also there is a common multiple between all \(  d_{i} \)