Hello!
Each normal distribution with a mean `mu` and standard deviation `sigma` may be made standard normal distribution by the linear transformation `x-gt(x-mu)/sigma,` where `x` is any value we are interesting in. Here `x_1 = 5.9` (hours) and `x_2 = 7.2` (hours).
Then we can use a table of probabilities for a standard normal distribution, I attached the link to it. Note that any normal distribution is symmetric, and the probability of "value less than the mean" is 0.5 (and 0.5, too, for "greater than the mean").
For using `x_1=5.9` with this table, we compute
`z_1 = (x_1-mu)/sigma = -0.3/1.1 approx -0.27` (- means to the left of the mean).
For `x_2=7.2` we have `z_2 = (x_2-mu)/sigma = 1/1.1 approx 0.91.`
The table values for `|z_1|` and `z_2` are approximately `0.11` and `0.32.` Now we can answer our questions.
1. Less than `7.2.` It is the left half of values with the probability 0.5, and those between 1/2 and `z_2,` for which we know the probability from the table. So the probability is about `0.5+0.32=0.82.`
2. Less than `5.9.` It is the left half of values minus those between `5.9` and `6.2,` i.e. the probability is about `0.5-0.11=0.39.`
`3` . Between `5.9` and `7.2.` It is "between `5.9` and `6.2`" plus "between `6.2` and `7.2`", i.e. about `0.11+0.32=0.43.`
We can use more precise table or online solving tool for the more precise results.
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