Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:b = 10 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Given that, mean of 10 observations is 10.
So, it means
Number of observations, n = 10
Mean of 10 observations, [tex] \overline{x}[/tex] = 10
We know,
Mean for n observations is given by
[tex]\sf \: \overline{x} \: = \: \dfrac{ \displaystyle\sum _{i = 1}^{n} \: x_i \: }{n} \\ \\ [/tex]
On substituting the values, we get
[tex]\sf \: 10 \: = \: \dfrac{ \displaystyle\sum _{i = 1}^{10} \: x_i \: }{10} \\ \\ [/tex]
[tex]\sf\implies \: \displaystyle\sum _{i = 1}^{10} \: x_i \: = \: 100 \\ \\ [/tex]
Now, while calculations, one observation 10 was left over.
So,
[tex]\sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: \displaystyle\sum _{i = 1}^{10} \: x_i \: + 10 \\ \\ [/tex]
[tex]\sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: 100\: + 10 \\ \\ [/tex]
[tex]\sf\implies \sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: 110\: \\ \\ [/tex]
Hence, the corrected mean is
[tex]\sf \: \overline{x} = \dfrac{\displaystyle\sum _{i = 1}^{11} \: x_i \: }{11} \\ \\ [/tex]
[tex]\sf \: \overline{x} = \dfrac{110}{11} \\ \\ [/tex]
[tex]\sf\implies \sf \: \overline{x} = 10 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
1. Mode of the continuous series is given by
[tex]{\boxed{{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}} \\ \\ [/tex]
where,
l is lower limit of modal class.
[tex] \sf{f_1} [/tex] is frequency of modal class
[tex] \sf{f_0} [/tex] is frequency of class preceding modal class
[tex] \sf{f_2} [/tex] is frequency of class succeeding modal class
h is class height.
2. Mean using Direct Method
[tex]\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ [/tex]
3. Mean using Short Cut Method
[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ [/tex]
4. Mean using Step Deviation Method
[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ [/tex]
mean is equal to 10.........
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Verified answer
Answer:
[tex]\qquad\qquad\qquad\boxed{ \bf{ \:b = 10 \: \: }} \\ \\ [/tex]
Step-by-step explanation:
Given that, mean of 10 observations is 10.
So, it means
Number of observations, n = 10
Mean of 10 observations, [tex] \overline{x}[/tex] = 10
We know,
Mean for n observations is given by
[tex]\sf \: \overline{x} \: = \: \dfrac{ \displaystyle\sum _{i = 1}^{n} \: x_i \: }{n} \\ \\ [/tex]
On substituting the values, we get
[tex]\sf \: 10 \: = \: \dfrac{ \displaystyle\sum _{i = 1}^{10} \: x_i \: }{10} \\ \\ [/tex]
[tex]\sf\implies \: \displaystyle\sum _{i = 1}^{10} \: x_i \: = \: 100 \\ \\ [/tex]
Now, while calculations, one observation 10 was left over.
So,
[tex]\sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: \displaystyle\sum _{i = 1}^{10} \: x_i \: + 10 \\ \\ [/tex]
[tex]\sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: 100\: + 10 \\ \\ [/tex]
[tex]\sf\implies \sf \: \displaystyle\sum _{i = 1}^{11} \: x_i \: = \: 110\: \\ \\ [/tex]
Hence, the corrected mean is
[tex]\sf \: \overline{x} = \dfrac{\displaystyle\sum _{i = 1}^{11} \: x_i \: }{11} \\ \\ [/tex]
[tex]\sf \: \overline{x} = \dfrac{110}{11} \\ \\ [/tex]
[tex]\sf\implies \sf \: \overline{x} = 10 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
1. Mode of the continuous series is given by
[tex]{\boxed{{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}} \\ \\ [/tex]
where,
l is lower limit of modal class.
[tex] \sf{f_1} [/tex] is frequency of modal class
[tex] \sf{f_0} [/tex] is frequency of class preceding modal class
[tex] \sf{f_2} [/tex] is frequency of class succeeding modal class
h is class height.
2. Mean using Direct Method
[tex]\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ [/tex]
3. Mean using Short Cut Method
[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ [/tex]
4. Mean using Step Deviation Method
[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ [/tex]
Step-by-step explanation:
mean is equal to 10.........