Question #47838

It is said that sufferers of a cold virus experience symptoms for 7 days. However, the amount of time is actually a normally distributed random variable whose mean is 7.5 days and whose standard deviation is 1.5 days.
A) What proportion of cold sufferers experience symptoms for less than 7.5 days?
B) What proportion of cold sufferers experience symptoms for between 7.5 and 10.5 days?
C) Compute the probability of a cold sufferer experiencing symptoms for more than 12 days.
D) Compute the probability of a cold sufferer experiencing symptoms for less than 5 days?
E) Compute the probability of a cold sufferer experiencing symptoms for between 6 and 8 days?
F) What cold length (number of days of symptoms) is exceeded by only 10% of cold sufferers?

Expert's answer

The normal distribution has probability density:

f(x) = 1/Sqrt[2 Pi]/sigma Exp[-(x -mean)^2/2/sigma^2)]

In our case

sigma = 1.5

mean = 7.5

a.& proportion of cold sufferers experiences less than 4 days of symptoms equals:

P(x<4) = Integrate[1/Sqrt[2 Pi]/1.5 Exp[-(x - 7.5)^2/2/1.5^2], {x, -Infinity, 4}] = 0.00176897 = 0.18 %

b.& proportion of cold sufferers experiences symptoms for between 7 and 10 days equals:

P(7<x<10) = Integrate[1/Sqrt[2 Pi]/1.5 Exp[-(x - 7.5)^2/2/1.5^2], {x, 7, 10}] = 0.642928 = 64.3 %

f(x) = 1/Sqrt[2 Pi]/sigma Exp[-(x -mean)^2/2/sigma^2)]

In our case

sigma = 1.5

mean = 7.5

a.& proportion of cold sufferers experiences less than 4 days of symptoms equals:

P(x<4) = Integrate[1/Sqrt[2 Pi]/1.5 Exp[-(x - 7.5)^2/2/1.5^2], {x, -Infinity, 4}] = 0.00176897 = 0.18 %

b.& proportion of cold sufferers experiences symptoms for between 7 and 10 days equals:

P(7<x<10) = Integrate[1/Sqrt[2 Pi]/1.5 Exp[-(x - 7.5)^2/2/1.5^2], {x, 7, 10}] = 0.642928 = 64.3 %

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