Probability of several consecutive drawn random samples of a standard normal distribution exceeding certain value.

Question

Hi,

I have trouble figuring out the following:

I have a standard normal distribution (µ = 0, σ² = 1) and now, as an example, draw 3 random independent samples.

What is the chance, that the sum of those samples is >0?

What is the chance, that the sum of those samples is >1?

What is the chance, that the sum of those samples is >3?

Answer 1

>0 is 50%

>1 is (1-.68)/2 = .32/2 = .16 = 16%

>3 is (1-.997)/2=.003/2=..0015 = 0.15%

half the population is above the mean, for 50%, half is below the mean

Use the 68-95-99.7 rule to calculate the rest

68% of the population is within 1 standard deviation of the mean, so 1-.68 is above and below that interval,

leaving (1-.68)/2 for the probability above 1 standard deviation.

99.7% fall within 3 standard deviations, so 1-.997 is above & below that interval, leaving half that above 3 standard deviations or (1-.997)/2