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

$\begin{array}{1 1}(A)\;81 \\(B)\;57 \\(C)\;36 \\(D)\;25 \end{array}$

Give the $SD = 5^{\circ}C$
Variance $=(S.D)^2$
$\qquad= (5^{\circ}C)^2$$=25^{\circ}C To convert ^{\circ}C to ^{\circ}F => \large\frac{^{\circ}C \times 9}{5}$$+32$
=> $\large\frac{25^{\circ} C \times 9}{5}$$+32$
=> $45 +32= 77^{\circ}F$
Hence A is the correct answer.