JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 16)
The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :
0
1
2
4
Explanation
Here mean = $$\overline x $$ = 9
$$ \Rightarrow $$ $$\overline x $$ = $${{\sum {{x_i}} } \over n}$$ = 9
$$ \Rightarrow $$ $${\sum {{x_i}} }$$ = 9 $$ \times $$ 5 = 45
Now, standard deviation = 0
$$\therefore\,\,\,$$ all the five terms are same i.e.; 9
Now for changed observation
$${\overline x _{new}}$$ = $${{36 + {x_5}} \over 5} = 10$$
$$ \Rightarrow $$ x5 = 14
$$\therefore\,\,\,$$ $$\sigma $$new = $$\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}} $$
= $$\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}} $$ = 2
$$ \Rightarrow $$ $$\overline x $$ = $${{\sum {{x_i}} } \over n}$$ = 9
$$ \Rightarrow $$ $${\sum {{x_i}} }$$ = 9 $$ \times $$ 5 = 45
Now, standard deviation = 0
$$\therefore\,\,\,$$ all the five terms are same i.e.; 9
Now for changed observation
$${\overline x _{new}}$$ = $${{36 + {x_5}} \over 5} = 10$$
$$ \Rightarrow $$ x5 = 14
$$\therefore\,\,\,$$ $$\sigma $$new = $$\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}} $$
= $$\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}} $$ = 2
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