Question

The standard deviation of some temperature data (in°C) is 5. If the data were converted into °F, then the variance would be

• A. 81
• B. 57
• C. 36
• D. 25

Answer 1

Answer: (A)

81

Answer 2

The relationship between temperature in Celsius (°C) and Fahrenheit (°F) is given by:

$F = \frac{9}{5}C + 32$

Let $X$ represent the temperature data in °C, and $Y$ represent the temperature data in °F. So, $Y = \frac{9}{5}X + 32$.

We are given that the standard deviation of the temperature data in °C is $\sigma_C = 5$. We need to find the variance of the temperature data in °F, which is $\text{Var}(Y) = \sigma_F^2$.

For a linear transformation of a random variable $X$ given by $Y = aX + b$, the variance of $Y$ is related to the variance of $X$ by the formula:

$\text{Var}(Y) = a^2 \text{Var}(X)$

In this case, $Y = \frac{9}{5}X + 32$, so $a = \frac{9}{5}$ and $b = 32$. The variance of $X$ (temperature in °C) is $\text{Var}(C) = \sigma_C^2 = 5^2 = 25$.

Using the formula for the variance of a linear transformation:

$\text{Var}(F) = \text{Var}(\frac{9}{5}C + 32)$ $\text{Var}(F) = (\frac{9}{5})^2 \text{Var}(C)$ $\text{Var}(F) = (\frac{81}{25}) \times 25$ $\text{Var}(F) = 81$

Alternatively, using standard deviation:

The standard deviation of $Y = aX + b$ is $\sigma_Y = |a|\sigma_X$. $\sigma_F = |\frac{9}{5}| \sigma_C = \frac{9}{5} \times 5 = 9$. The variance is the square of the standard deviation: $\text{Var}(F) = \sigma_F^2 = 9^2 = 81$.

The variance of the temperature data in °F is 81.