Flashcards · 4040 · Topic 6

Transformations involving mean and standard deviation flashcards

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What happens to the mean and standard deviation when a constant a is added to every value?: The mean increases by a. The standard deviation (and variance) are unchanged, because the gaps between the values and the mean do not change.
What happens when every value is multiplied by a constant b?: Mean → b × mean. Standard deviation → |b| × standard deviation. Variance → b² × variance.
Times have mean 25 min and standard deviation 6 min. The charge is y = 10 + 4x. Find the mean and SD of y: Mean = 10 + 4 × 25 = 110. Standard deviation = 4 × 6 = 24.
x has mean 8 and standard deviation 1.5. Find the mean and standard deviation of y = 50 − 3x: Mean = 50 − 3 × 8 = 26. Standard deviation = |−3| × 1.5 = 4.5. A standard deviation is never negative.
Coded values y = (x − 80) ÷ 4 have mean 2.5 and SD 0.5. Find the mean and SD of x: Undo the coding: x = 80 + 4y. Mean of x = 80 + 4 × 2.5 = 90. SD of x = 4 × 0.5 = 2.
Does adding 10 to every observation change the range?: No. The largest and smallest values both rise by 10, so the range is unchanged, like the standard deviation.

How do you find y = a + bx that gives a target mean and standard deviation?: Find b first: b = target SD ÷ old SD. Then a = target mean − b × old mean. Check by substituting the old mean: it must give the target mean.
Data with mean 55 and SD 8 are to have mean 100 and SD 20. Find the transformation: b = 20 ÷ 8 = 2.5. a = 100 − 2.5 × 55 = −37.5. So y = −37.5 + 2.5x. Check: −37.5 + 137.5 = 100.
Data with mean 20 and SD 5 are to have mean 50 and SD 10. Find y = a + bx: b = 10 ÷ 5 = 2. a = 50 − 2 × 20 = 10. So y = 10 + 2x.
Data with mean 12 and SD 3 are to have mean 0 and SD 1. Find a and b: b = 1/3 and a = 0 − (1/3) × 12 = −4, so y = −4 + x/3. This is standardising the data.