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C# : is there an appropriate collection for fast range-related search?

I have data like that:
Time(seconds from start)
Value
15
2
16
4
19
2
25
9
There are a lot of entries (10000+), and I need a way to find fast enough sum of any time range, like sum of range 16-25 seconds (which would be 4+2+9=15). This data will be dynamically changed many times (always adding new entries at the bottom of list).
I am thinking about using sorted list + binary search to determinate positions and just make sum of values, but is can took too much time to calculate it. Is there are any more appropriate way to do so? Nuget packets or algorithm references would be appreciated.

Just calculate cumulative sum:
Time Value CumulativeSum
15 2 2
16 4 6
19 2 8
25 9 17
Then for range [16,25] it will be task to binary search left border of 16 and 25 exact, which turns into 17 - 2 = 15
Complexity: O(log(n)), where n - size of the list.
Binary search implementation for lower/upper bound can be found in my repo - https://github.com/eocron/Algorithm/blob/master/Algorithm/Sorted/BinarySearchExtensions.cs

How is an integer stored in memory?

This is most probably the dumbest question anyone would ask, but regardless I hope I will find a clear answer for this.
My question is - How is an integer stored in computer memory?
In c# an integer is of size 32 bit. MSDN says we can store numbers from -2,147,483,648 to 2,147,483,647 inside an integer variable.
As per my understanding a bit can store only 2 values i.e 0 & 1. If I can store only 0 or 1 in a bit, how will I be able to store numbers 2 to 9 inside a bit?
More precisely, say I have this code int x = 5; How will this be represented in memory or in other words how is 5 converted into 0's and 1's, and what is the convention behind it?

It's represented in binary (base 2). Read more about number bases. In base 2 you only need 2 different symbols to represent a number. We usually use the symbols 0 and 1. In our usual base we use 10 different symbols to represent all the numbers, 0, 1, 2, ... 8, and 9.
For comparison, think about a number that doesn't fit in our usual system. Like 14. We don't have a symbol for 14, so how to we represent it? Easy, we just combine two of our symbols 1 and 4. 14 in base 10 means 1*10^1 + 4*10^0.
1110 in base 2 (binary) means 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0 = 8 + 4 + 2 + 0 = 14. So despite not having enough symbols in either base to represent 14 with a single symbol, we can still represent it in both bases.
In another commonly used base, base 16, which is also known as hexadecimal, we have enough symbols to represent 14 using only one of them. You'll usually see 14 written using the symbol e in hexadecimal.
For negative integers we use a convenient representation called twos-complement which is the complement (all 1s flipped to 0 and all 0s flipped to 1s) with one added to it.
There are two main reasons this is so convenient:
We know immediately if a number is positive of negative by looking at a single bit, the most significant bit out of the 32 we use.
It's mathematically correct in that x - y = x + -y using regular addition the same way you learnt in grade school. This means that processors don't need to do anything special to implement subtraction if they already have addition. They can simply find the twos-complement of y (recall, flip the bits and add one) and then add x and y using the addition circuit they already have, rather than having a special circuit for subtraction.

This is not a dumb question at all.
Let's start with uint because it's slightly easier. The convention is:
You have 32 bits in a uint. Each bit is assigned a number ranging from 0 to 31. By convention the rightmost bit is 0 and the leftmost bit is 31.
Take each bit number and raise 2 to that power, and then multiply it by the value of the bit. So if bit number three is one, that's 1 x 23. If bit number twelve is zero, that's 0 x 212.
Add up all those numbers. That's the value.
So five would be 00000000000000000000000000000101, because 5 = 1 x 20 + 0 x 21 + 1 x 22 + ... the rest are all zero.
That's a uint. The convention for ints is:
Compute the value as a uint.
If the value is greater than or equal to 0 and strictly less than 231 then you're done. The int and uint values are the same.
Otherwise, subtract 232 from the uint value and that's the int value.
This might seem like an odd convention. We use it because it turns out that it is easy to build chips that perform arithmetic in this format extremely quickly.

Binary works as follows (as your 32 bits).
1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1
2^ 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16......................................0
x
x = sign bit (if 1 then negative number if 0 then positive)
So the highest number is 0111111111............1 (all ones except the negative bit), which is 2^30 + 2 ^29 + 2^28 +........+2^1 + 2^0 or 2,147,483,647.
The lowest is 1000000.........0, meaning -2^31 or -2147483648.

Is this what high level languages lead to!? Eeek!
As other people have said it's a base 2 counting system. Humans are naturally base 10 counters mostly, though time for some reason is base 60, and 6 x 9 = 42 in base 13. Alan Turing was apparently adept at base 17 mental arithmetic.
Computers operate in base 2 because it's easy for the electronics to be either on or off - representing 1 and 0 which is all you need for base 2. You could build the electronics in such a way that it was on, off or somewhere in between. That'd be 3 states, allowing you to do tertiary math (as opposed to binary math). However the reliability is reduced because it's harder to tell the difference between those three states, and the electronics is much more complicated. Even more levels leads to worse reliability.
Despite that it is done in multi level cell flash memory. In these each memory cell represents on, off and a number of intermediate values. This improves the capacity (each cell can store several bits), but it is bad news for reliability. This sort of chip is used in solid state drives, and these operate on the very edge of total unreliability in order to maximise capacity.

Generating random numbers for solar energy harvesting using Markov models

How do I generate random numbers using a Markov model in C#? I noticed here that almost all of the applications of the Markov algorithm is for randomly writing text. Is there a source code somewhere or a tutorial where I can fully understand how this works? My goal actually is to generate random numbers to simulate solar energy harvesting.

First decide how deep your Markov model is going. Do you look at the previous number? The previous two numbers? The previous three numbers? Perhaps deeper?
Second, look through some actual solar energy data and extract the probabilities for what follows a group of 1, 2 or 3 numbers. Ideally you will be able to get complete coverage, but there may well be gaps. For those either extrapolate, or put in some average/random value.
All this so far is data.
Third generate the first 1, 2 or 3 numbers. From your database pick the correct combination and randomly select one of the possible followers. When I do this, I have a low probability random element possible as well so things don't get stuck in a rut.
Drop the earliest element of your 1, 2 or 3 numbers. Shift the others down and add the new number at the end. Repeat until you have enough data.
Here is a short extract from my 1-deep Markov word generator showing part of the data table:
// The line addEntry('h', "e 50 a 23 i 12 o 7 # 100") shows the the letter
// 'h' is followed by 'e' 50% of the time, 'a' 23% of the time, 'i' 12% of
// the time, 'o' 7% of the time and otherwise some other letter, '#'.
//
// Figures are taken from Gaines and tweaked. (see 'q')
private void initMarkovTable() {
mMarkovTable = new HashMap<Character, List<CFPair>>(26);
addEntry('a', "n 21 t 17 s 12 r 10 l 8 d 5 c 4 m 4 # 100");
addEntry('b', "e 34 l 17 u 11 o 9 a 7 y 5 b 4 r 4 # 100");
addEntry('c', "h 19 o 19 e 17 a 13 i 7 t 6 r 4 l 4 k 4 # 100");
addEntry('d', "e 16 i 14 a 14 o 10 y 8 s 6 u 5 # 100");
addEntry('e', "r 15 d 10 s 9 n 8 a 7 t 6 m 5 e 4 c 4 o 4 w 4 # 100");
addEntry('f', "t 22 o 21 e 10 i 9 a 7 r 5 f 5 u 4 # 100");
addEntry('g', "e 14 h 14 o 12 r 10 a 8 t 6 f 5 w 4 i 4 s 4 # 100");
addEntry('h', "e 50 a 23 i 12 o 7 # 100");
// ...
}
The data is organised as letter-frequency pairs. I used the '#' character to indicate "pick any letter here". Your data will be number-frequency pairs instead.
To pick an output, I read the appropriate line of the data and generate a random percentage. Scan long the data accumulating the frequencies until the accumulated frequency exceeds the random percantage. That is the letter (or number in your case) you pick.

BitArray returns bits the wrong way around?

This code:
BitArray bits = new BitArray(new byte[] { 7 });
foreach (bool bit in bits)
{
Console.WriteLine(bit ? 1 : 0);
}
Gives me the following output:
11100000
Shouldn't it be the other way around? Like this:
00000111
I am aware that there is little and big endian, although those terms only refer to the position of bytes. As far as I know, they don't affect bits.

The documentation for BitArray states:
The first byte in the array represents bits 0 through 7, the second
byte represents bits 8 through 15, and so on. The Least Significant
Bit of each byte represents the lowest index value: " bytes [0] & 1"
represents bit 0, " bytes [0] & 2" represents bit 1, " bytes [0] & 4"
represents bit 2, and so on.
When indexing bits, the convention is to start at the least significant end, which is the right side when written in binary notation. However, when enumerating the array, you start at index 0, so they are printed out left-to-right instead of right-to-left. That's why it looks backwards.
For example, the word 01011010 00101101 (90 45) would be indexed as:
0 1 0 1 1 0 1 0 - 0 0 1 0 1 1 0 1
----------------------- -----------------------
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
And you would pass it to the constructor as new byte[] { 45, 90 } since you pass it least-significant first. When printed out, it would display in index order as: 1011010001011010, which is the reverse of the original binary notation.

The documentation is not explicit about it, but I guess the iterator iterates from the LSB to the MSB. Sound reasonable to me (personally!), unless you are printing out the bits. I had a look at BitArray.GetEnumerator Method.

No it's a bit array, not a numeric value represented as bits.
It's just like any regular array with some methods added for bit operations. Just like if you had an array of int. You wouldn't expect it to be in the reverse order, it just takes it position by position.
Eg:
Numbers (in a byte) converted to a BitArray would come out like:
2 = 01000000
5 = 10100000
8 = 00010000
etc.
It just stores the position of the value but not relative as you would except from a binary numeric value.
Here is a link describing the constructor you are using:
http://msdn.microsoft.com/en-us/library/b3d1dwck.aspx
The key point is:
The number in the first values array element represents bits 0 through
31, the second number in the array represents bits 32 through 63, and
so on. The Least Significant Bit of each integer represents the lowest
index value: " values [0] & 1" represents bit 0, " values [0] & 2"
represents bit 1, " values [0] & 4" represents bit 2, and so on.

small problem in shifting range please help me out

I have small concept problem.
i am working on graph.
consider i hav a point 32 between range 28 and 35
and i have to bring all the points between the range 28 and 35 within range 1 and 2.
how to calculate it?
actually i ll have the point 32.
and i have to shift it between 1 and 2...
please help me out.
In other words,
if 32 is between 28 and 35
what is 32 in range 1 and 2

I think it is: 1 + [(number - 28) / (35 - 28)], for example for 32 is (1 + 4/7) = 1.57...
and in general if you want move it within [a,b]:
a + (b-a) * [(number - 28) / (35 - 28)]