Answer:
a) 

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %
b) 

So one deviation below the mean we have: (100-68)/2 = 16%
c) 

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%
Step-by-step explanation:
For this case we have a random variable with the following parameters:

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.
We want to find the following probability:

We can find the number of deviation from the mean with the z score formula:

And replacing we got


And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %
For the second case:


So one deviation below the mean we have: (100-68)/2 = 16%
For the third case:

And replacing we got:


For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%
Answer:
u = 16
Step-by-step explanation:
Given
u = 12
Multiply both sides by 4 to eliminate the fraction
3u = 48 ( divide both sides by 3 )
u = 16
Answer:
Step-by-step explanation:
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