
A student scores the following marks in five tests: $45,54,41,57,43$. His score is not known for the sixth test. If the mean score is 48 for the six tests, then what is the standard deviation of the marks in six tests?
A) $\dfrac{{10}}{{\sqrt 3 }}$
B) $\dfrac{{100}}{{\sqrt 3 }}$
C) $\dfrac{{130}}{3}$
D) $\dfrac{{10}}{3}$
Answer
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Hint: In this question we are given the marks of the student in five tests along with the mean of six tests. We have been asked to find the standard deviation of the marks in six tests. To find standard deviation, first find the sixth observation using the formula of mean. After that, each observation is squared and added to put in the formula of variance. Now, we have all the values. Put them in the formula and find variance. Next, square root the variance to find standard deviation.
Formula used: 1) Standard deviation $ = \sqrt {{\sigma ^2}} = \sqrt {{\text{Variance}}} $
2) Variance ${(\sigma )^2} = \left( {\dfrac{{\sum\limits_{i = 1}^6 {{x_i}^2} }}{N} - {{\left( {\bar X} \right)}^2}} \right)$
Complete step-by-step solution:
We are given the scores of 5 tests and a mean of 6 tests. Then we are asked to find the standard deviation of the marks in six tests.
At first, we will find the marks in the sixth test using the 5 scores and the given mean. Let the 6th observation be x.
$\bar X = \dfrac{{{\text{sum of observations}}}}{{{\text{total observations}}}}$
Putting all the values,
$ \Rightarrow 48 = \dfrac{{45 + 54 + 41 + 57 + 43 + x}}{6}$
Solving for x,
$ \Rightarrow 48 \times 6 = 240 + x$
$ \Rightarrow 288 - 240 = x$
$ \Rightarrow x = 48$
Now, we have all the six observations.
Now we know that standard deviation = $\sqrt {{\text{Variance}}} $
Variance${(\sigma )^2}$ = $\left( {\dfrac{{\sum\limits_{i = 1}^6 {{x_i}^2} }}{N} - {{\left( {\bar X} \right)}^2}} \right)$
Putting all the values in the formula,
$ \Rightarrow {\sigma ^2} = \left( {\dfrac{{({{45}^2} + {{54}^2} + {{41}^2} + {{57}^2} + {{43}^2} + {{48}^2})}}{6} - {{\left( {48} \right)}^2}} \right)$
Simplifying we get,
$ \Rightarrow {\sigma ^2} = \left( {\dfrac{{14024}}{6} - 2304} \right)$
$ \Rightarrow {\sigma ^2} = 33.33 = \dfrac{{100}}{3}$
Now, standard deviation = $\sqrt {{\text{Variance}}} $
$ \Rightarrow \sigma = \sqrt {\dfrac{{100}}{3}} = \dfrac{{10}}{{\sqrt 3 }}$
$\therefore $ The standard deviation of the marks in six tests = $\dfrac{{10}}{{\sqrt 3 }}$.
Option A is the correct answer.
Note: The Standard deviation is a statistic that looks at how far from the mean a group of numbers is, by using the square root of the variance. The calculation of the variance used squares because it weighs outliers more heavily than data closer to the mean. This calculation also prevents differences above the mean from cancelling out those below, which would result in a variance of zero.
Formula used: 1) Standard deviation $ = \sqrt {{\sigma ^2}} = \sqrt {{\text{Variance}}} $
2) Variance ${(\sigma )^2} = \left( {\dfrac{{\sum\limits_{i = 1}^6 {{x_i}^2} }}{N} - {{\left( {\bar X} \right)}^2}} \right)$
Complete step-by-step solution:
We are given the scores of 5 tests and a mean of 6 tests. Then we are asked to find the standard deviation of the marks in six tests.
At first, we will find the marks in the sixth test using the 5 scores and the given mean. Let the 6th observation be x.
$\bar X = \dfrac{{{\text{sum of observations}}}}{{{\text{total observations}}}}$
Putting all the values,
$ \Rightarrow 48 = \dfrac{{45 + 54 + 41 + 57 + 43 + x}}{6}$
Solving for x,
$ \Rightarrow 48 \times 6 = 240 + x$
$ \Rightarrow 288 - 240 = x$
$ \Rightarrow x = 48$
Now, we have all the six observations.
Now we know that standard deviation = $\sqrt {{\text{Variance}}} $
Variance${(\sigma )^2}$ = $\left( {\dfrac{{\sum\limits_{i = 1}^6 {{x_i}^2} }}{N} - {{\left( {\bar X} \right)}^2}} \right)$
Putting all the values in the formula,
$ \Rightarrow {\sigma ^2} = \left( {\dfrac{{({{45}^2} + {{54}^2} + {{41}^2} + {{57}^2} + {{43}^2} + {{48}^2})}}{6} - {{\left( {48} \right)}^2}} \right)$
Simplifying we get,
$ \Rightarrow {\sigma ^2} = \left( {\dfrac{{14024}}{6} - 2304} \right)$
$ \Rightarrow {\sigma ^2} = 33.33 = \dfrac{{100}}{3}$
Now, standard deviation = $\sqrt {{\text{Variance}}} $
$ \Rightarrow \sigma = \sqrt {\dfrac{{100}}{3}} = \dfrac{{10}}{{\sqrt 3 }}$
$\therefore $ The standard deviation of the marks in six tests = $\dfrac{{10}}{{\sqrt 3 }}$.
Option A is the correct answer.
Note: The Standard deviation is a statistic that looks at how far from the mean a group of numbers is, by using the square root of the variance. The calculation of the variance used squares because it weighs outliers more heavily than data closer to the mean. This calculation also prevents differences above the mean from cancelling out those below, which would result in a variance of zero.
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