Answer:
a) The expected value is given by:

The variance is given by:

And the standard deviation is just the square root of the variance and we got:

b) 
And we can use the cumulative distribution function given by:

And using this function we got:

c) 
And using the complement rule and the cumulative distribution function we got:

Step-by-step explanation:
For this case we define the random variable X=quantity put in each bag , and we know that the distribution for X is given by:

Part a
The expected value is given by:

The variance is given by:

And the standard deviation is just the square root of the variance and we got:

Part b
For this case we want this probability:

And we can use the cumulative distribution function given by:

And using this function we got:

Part c
For this case we want this probability:

And using the complement rule and the cumulative distribution function we got:
