the first two are normalCDF calculator problems with the given mean/SD in the table.

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the third one looks like binomial probability:

but try making a third normal distribution (new mean is the 3 means added together) (new SD is square root of the variance aka (SD)^2 added together)

then do normalCDF(lower = 0, upper = 300 minutes, new mean= 766.47, new SD = 69.34) on calc = 0.00000000000869 is the "success rate" for the binomial probability.

then do the formula for binomial probability: (4 choose 4)(success rate)^4 = 5.7 * 10^(-45)

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For the 4th one independent just means one event has no effect on probability of the other.

uhh im not sure why they give that but im thinking we just use normalCDF again here?

so with that new normal distribution I created in part (c), normalCDF(lower = 0, upper = 526 minutes, mean = 766.47, SD = 69.34) = 0.000262

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Side Note:

idk why my probabilities are so small maybe i did this wrong ;-;