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NCERT Exemplar · Q39

Q.The standard deviation of some temperature data in °C is 5. If the data were converted into ºF, the variance would be
(A) 81
(B) 57
(C) 36
(D) 25

Uttarakhand UbseMCQ· 1mImportance★★★★★est
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When data is transformed linearly, Y=aX+bY = aX + b, the variance scales by a2a^2 and is unaffected by bb. Converting Celsius to Fahrenheit involves a scaling factor of 95\frac{9}{5}, so the variance in Fahrenheit will be (95)2(\frac{9}{5})^2 times the variance in Celsius. Given a standard deviation of 5°C, the variance in °F is 81\boxed{81}.

When we talk about the spread or variability of data, like standard deviation or variance, we are measuring how much the individual data points deviate from the mean. If we apply a linear transformation to the data, say Y=aX+bY = aX + b, where aa and bb are constants, we need to understand how this affects the spread.

Consider the effect of the additive constant bb. If we add a constant value to every data point, the entire dataset shifts up or down by that constant amount. The mean also shifts by the same amount. However, the distances between data points remain unchanged. For example, if we have data points 1, 2, 3, and we add 10 to each, we get 11, 12, 13. The difference between 1 and 2 is 1, and the difference between 11 and 12 is also 1. Since standard deviation and variance are based on these differences (specifically, deviations from the mean), adding a constant bb does not change them.

Now, consider the effect of the multiplicative constant aa. If we multiply every data point by a constant aa, the distances between data points also get scaled by ∣a∣|a|. For instance, if we have 1, 2, 3 and multiply by 2, we get 2, 4, 6. The original difference of 1 (between 1 and 2) becomes 2 (between 2 and 4). So, the standard deviation will be scaled by ∣a∣|a|. Since variance is the square of the standard deviation, it will be scaled by a2a^2.

For a linear transformation Y=aX+bY = aX + b:

Var(Y)=a2Var(X)Var(Y) = a^2 Var(X)

SD(Y)=∣a∣SD(X)SD(Y) = |a| SD(X)

Let's apply this understanding to the problem.

  1. Identify the given information:

    We are given the standard deviation of temperature data in °C:

    SD(C)=5SD(C) = 5

  2. Recall the conversion formula from Celsius to Fahrenheit:

    The formula to convert temperature from Celsius (CC) to Fahrenheit (FF) is:

    F=95C+32F = \frac{9}{5}C + 32

  3. Identify the scaling and shifting constants:

    Comparing this to the general linear transformation Y=aX+bY = aX + b, we have: …

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