
Let ${{x}_{i}}(1\le i\le 10)$be ten observations of a random variable X. if
$\sum\limits_{i=1}^{10}{({{x}_{i}}-p)}$ = 3
And $\sum\limits_{i=1}^{10}{{{({{x}_{i}}-p)}^{2}}}$ = 9
Where 0 $\ne p\in R$ then the standard deviation of these observations is
(a)$\dfrac{7}{10}$
(b)$\dfrac{9}{10}$
( c)$\sqrt{\dfrac{3}{5}}$
(d)$\dfrac{4}{5}$
Answer
164.1k+ views
Hint: We are given the ten observations of a random variable and we have to find out the standard observation. To solve this question, first we find out the mean of the given observations. Then we find out the variance but they already give us these values. So we do not need to find it. We use the formula $\dfrac{\sum{{{({{x}_{i}}-p)}^{2}}}}{n}$ - ${{\left( \dfrac{\sum{({{x}_{i}}-p)}}{n} \right)}^{2}}$. We simply subtract the values and get our desirable answer.
Complete step by step solution:
Given that
$\sum\limits_{i=1}^{10}{({{x}_{i}}-p)}$ = 3
And $\sum\limits_{i=1}^{10}{{{({{x}_{i}}-p)}^{2}}}$ = 9
As the standard deviation is free from shifting of origin.
As the values of variance are given. We directly use these values to find out the answer.
We know the formula of variance = $\dfrac{\sum{{{({{x}_{i}}-p)}^{2}}}}{n}$ - ${{\left( \dfrac{\sum{({{x}_{i}}-p)}}{n} \right)}^{2}}$
Standard deviation = $\sqrt{\operatorname{var}iance}$
= $\sqrt{\left[ \left( \dfrac{9}{10} \right)-{{\left( \dfrac{3}{10} \right)}^{2}} \right]}$
By solving the above equation, we get
$\sqrt{\left[ \left( \dfrac{9}{10} \right)-{{\left( \dfrac{3}{10} \right)}^{2}} \right]}$ = $\sqrt{\left[ \left( \dfrac{9}{10} \right)-\left( \dfrac{9}{100} \right) \right]}$
By solving it, we get
$\sqrt{\left[ \left( \dfrac{9}{10} \right)-\left( \dfrac{9}{100} \right) \right]}$ = $\sqrt{\left[ \left( \dfrac{81}{100} \right) \right]}$
Then the value of $\sqrt{\left[ \left( \dfrac{81}{100} \right) \right]}$ = $\dfrac{9}{10}$
Hence the value of standard observations of these variables is $\dfrac{9}{10}$
Thus, Option (B) is correct.
Note: In these types of questions, students made mistakes in the calculation part. As the values of variance are given some students get confused and they try to find out the variance which is already given in the question. In this question, mainly students have to know the formula of variance, then the student will be able to find out the answer easily. Be careful while doing calculations as many students made mistakes in the calculation part
Complete step by step solution:
Given that
$\sum\limits_{i=1}^{10}{({{x}_{i}}-p)}$ = 3
And $\sum\limits_{i=1}^{10}{{{({{x}_{i}}-p)}^{2}}}$ = 9
As the standard deviation is free from shifting of origin.
As the values of variance are given. We directly use these values to find out the answer.
We know the formula of variance = $\dfrac{\sum{{{({{x}_{i}}-p)}^{2}}}}{n}$ - ${{\left( \dfrac{\sum{({{x}_{i}}-p)}}{n} \right)}^{2}}$
Standard deviation = $\sqrt{\operatorname{var}iance}$
= $\sqrt{\left[ \left( \dfrac{9}{10} \right)-{{\left( \dfrac{3}{10} \right)}^{2}} \right]}$
By solving the above equation, we get
$\sqrt{\left[ \left( \dfrac{9}{10} \right)-{{\left( \dfrac{3}{10} \right)}^{2}} \right]}$ = $\sqrt{\left[ \left( \dfrac{9}{10} \right)-\left( \dfrac{9}{100} \right) \right]}$
By solving it, we get
$\sqrt{\left[ \left( \dfrac{9}{10} \right)-\left( \dfrac{9}{100} \right) \right]}$ = $\sqrt{\left[ \left( \dfrac{81}{100} \right) \right]}$
Then the value of $\sqrt{\left[ \left( \dfrac{81}{100} \right) \right]}$ = $\dfrac{9}{10}$
Hence the value of standard observations of these variables is $\dfrac{9}{10}$
Thus, Option (B) is correct.
Note: In these types of questions, students made mistakes in the calculation part. As the values of variance are given some students get confused and they try to find out the variance which is already given in the question. In this question, mainly students have to know the formula of variance, then the student will be able to find out the answer easily. Be careful while doing calculations as many students made mistakes in the calculation part
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