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
.
A very small dinosaur , the microraptor was only 1.3 feet long. one of the largest dinosaures the diplodocus was about 91 feet long. how many times as long as the microraptor was the diplodocus ? =78 feet
Answer:
-5.02
Step-by-step explanation:
the result becomes negative because 7.273 is less than 12.37
Answer:
elaborate, this is vague
Step-by-step explanation:
Answer:
you need 5 buses and 4 will be totally full and one will only be 1/3 full
or you can have 14% of all 5 buses with empty seats
Step-by-step explanation: