Morning temperature above a threshold

Question

In a certain month, the dawn temperature is equally likely to be anywhere between $-5\,^\circ$ C and $15\,^\circ$ C. That is, the temperature is uniform across a 20-degree range.

What is the probability that the temperature is above $9\,^\circ$ C?

Accepted Answer

For a value spread evenly over an interval, the probability of a target region is its width divided by the width of the whole range. P(	ext{event}) = \frac{	ext{width of favorable region}}{	ext{width of full range}}. The full range runs from -5 to 15, a width of 15 - (-5) = 20 degrees. "Above 9\,^\circC" means the stretch from 9 up to 15, which is 15 - 9 = 6 degrees wide. So P(	ext{temp} 9) = \frac{6}{20} = \frac{3}{10} = 0.3. Be careful with the negative endpoint: the total width is 20, not 15. Subtracting the smaller endpoint (here -5) is what keeps the arithmetic honest. Takeaway. Watc…

Morning temperature above a threshold

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