The expected length of code for one encoded symbol is

where
is the probability of picking the letter
, and
is the length of code needed to encode
.
is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length
, we would then expect
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length
, we would get
.
100%-30%=70%
100%+30%=130%
to decrese by 30%, multiply by 70%
to increase by 30%, multiply by 130%
Based upon the answer choices, the answer is A;
A Straight line passing through the origin.
I hope this answer has assisted you.
Answer:
eight away .7 can be used into 3 4 times with a sum of 2.8 and that leaves .2 which would be left out since it is not a whole .7 so C
Step-by-step explanation: