Question

You observe the content of 2 email messages. An email may contain an image with probability 0.35. An image is equally likely to be 3 MB or 6 MB. Assume that an email without an image is 1 MB.

a) Determine the probability mass function of X (the total size of all emails observed)

b) What is the expected value of the total size of all emails?

c) What is the standard deviation of the total size of all emails.

Answer 1

a) An email may contain an image with probability 0.35.

The probability that email contains no image = 1-0.35 = 0.65

An image is equally likely to be 3 MB or 6 MB.

Therefore, the probability that email contains an image of 3 MB = 0.35/2 = 0.175

and also the probability that email contains an image of 6 MB = 0.35/2 = 0.175

The probability mass function is:

x (size in MB)	P(x)
1	0.650
3	0.175
6	0.175
Sum	1.00

----------------------------------------------------------------------------------------------

b) Consider the following table:

x	P(x)	x P(x)	x²P(x)
1	0.650	0.650	0.650
3	0.175	0.525	1.575
6	0.175	1.050	6.300
Sum	1.000	2.225	8.525

The expected value of the total size of all emails is:

----------------------------------------------------------------------------------------------

c) The standard deviation of the total size of all emails is: