Question

The standard deviation of some temperature data in ${}^{\circ }C$ is $5$ . if the data were converted into ${}^{\circ }F$ , the variance would be

Answer 1

From the question given that we have to find the variance for the standard deviation of some temperature ${}^{\circ }C$ is $5$ and data has been converted into ${}^{\circ }F$. As we know that the relation between the ${}^{\circ }F$ and ${}^{\circ }C$ is ${}^{\circ }F=\dfrac{9C}{5}+32$. As we know that if standard deviation of x-series is s, then the standard deviation of Kx-series is Ks, and the standard deviation of $k+x$series is s. by squaring the standard deviation we will get the variance.

Complete step by step solution:
From the question given that the standard deviation of some temperature ${}^{\circ }C$ is
$\Rightarrow {{\sigma }_{C}}=5$
Now, to convert the data into the ${}^{\circ }F$, As we know that the relation between the ${}^{\circ }F$ and ${}^{\circ }C$ is
$\Rightarrow {}^{\circ }F=\dfrac{9C}{5}+32$
As we know that if the standard deviation of x-series is s, then the standard deviation of Kx-series is Ks, and the standard deviation of $k+x$ series is s.
$\Rightarrow {{\sigma }_{{}^{\circ }F}}=\dfrac{9{{\sigma }_{C}}}{5}+{{\sigma }_{32}}$
Now we have to substitute the values in their respective positions,
By substituting we will get,
$\Rightarrow {{\sigma }_{{}^{\circ }F}}=\dfrac{9\times 5}{5}+0$
By simplifying further, we will get,
$\Rightarrow {{\sigma }_{{}^{\circ }F}}=9$
Therefore, as we know that the square of the standard deviation is equal to the variance, that is
$\Rightarrow \text{variance}\left( X \right)={{\sigma }^{2}}$
Now, squaring on both sides we will get,
$\Rightarrow {{\left( {{\sigma }_{{}^{\circ }F}} \right)}^{2}}={{\left( 9 \right)}^{2}}$
$\Rightarrow {{\left( {{\sigma }_{{}^{\circ }F}} \right)}^{2}}=81$

Therefore, after converting the data into ${}^{\circ }F$, the variance is equal to the $81$.

Note: Students should know the formulas of statistics and students should not confuse about the relation between standard deviation and variance, if students write $\text{variance}\left( X \right)=\sqrt{\sigma }$ instead of $\Rightarrow \text{variance}\left( X \right)={{\sigma }^{2}}$ , the whole answer will be wrong.