What is the frequency of heterozygotes for this population?

Thus, the present year population ought to be around 19.8075 Million and the anticipated population of New York of the year 2016 ought to be 19.865 Million. The US Bureau estimates that the population of New Yorkwas 19,746,227 on July 1, 2014, a 1.9% increase since the 2010 United States Census.17 Jan 2016

Population Of New York In 2016

population2016.com/population-of-new-york-in-2016.html

About 19.7

Population size:4
Mean (?): 23.45
Standard deviation (?): 4.5986411036305

In the 2010 census, Sulphur Springs, Indiana had a population of 399 and Sulphur Springs, Texas had a population of 15,449.

The population doubles in ten years so the annual growth is the tenth root of two. The growth in twelve years is the annual growth rate raised to the twelfth power.

The tenth root of two is about 1.0718 . That value raised to the twelfth power is about 2.2974 . Multiplied by the initial population of 2800 gives 6433.

On the TI-30XA press 2 2nd [xth-root] 1 0 = y^x 1 2 * 2 8 0 0 =

Example: Enter the following x variables, and determine the sample and population standard deviations.
x = {1,2,3}

• Press [DATA] and begin to enter the data as indicated into the L1, L2 or L3 lists. To input a data point, type the value and press [ENTER]. Press the down arrow key to move on to the next data point.
• Press [2nd], [STAT], and then choose 1-VAR.
• For the "DATA" option, choose the listname the data is stored in (L1, L2 or L3). For the "FRQ" option, choose "ONE" since this data does not include a frequency list.
• Arrow down to "CALC" and press [ENTER].
  
  Answers should be as follows:
  3:Sx (Sample Standard Deviation) = 1
  4:sx (Population Standard Deviation) = 0.816497
• Press [2nd] [MODE] to exit STAT mode.

Hello,
I am afraid I do not understand. The frequency of a PARTICULAR score is the number of times that particular value occurs in the data. It is not the number of data.
Ex: the data are:
4, 6, 5, 4, 4, 8, 9, 3, 3, 6, 7
Frequency of (3) =2
Frequency of (4)= 3
Frequency of (5) =1
Frequency of (6)= 2
Frequency of (7)=1
Frequency of (8)=1
Frequency of (9)=1

Number of data =11 (n=11)

You compute the mean M
M= (2x3 + 3x4 +1x5 +2x6+1x7+1x8+1x9) /11 =5.363636..
As you can see, if a term is not repeated it is multiplied by 1 (its frequency is one). Value 3 occurs twice (hence 2x3); value 4 occurs 3 times (3x4), etc.

To calculate by hand the sum of squares
3: (3-5.3636)^2 + (3-5.3636)^2 .................... = 2x(3-5.3636)^2
4: (4-5.3636)^2 + (4-5.3636)^2 +(4-5.3636)^2 = 3x(4-5.3636)^2
5: (5-5.3636)^2 ...................................... .= 1x(5-5.3636)^2
6: (6-5.3636)^2 +(6-5.3636)^2......................= 2x(6-5.3636)^2
7: (7-5.3636)^2.......................................... .=1x(7-5.3636)^2
8: (8-5.3636)^2......................................... ..=1x(8-5.3636)^2
9: (9-5.3636)^2......................................... ..=1x(9-5.3636)^2
If I have not made a mistake the sum of squares is 40.5454

Standard Deviation

The standard formula above gives s= square root (40.5454/10) =2.01
Population Standard Deviation

The population Standard deviation above is S= square root (40.5454/11) =1.9198

So if you perform the calculation with the calculator the only times you need to enter the frequency is for repeated terms. When you have to enter 6 above, its frequency is 2 you proceed as follows
6 [2nd][FRQ] 2 [Sigma+]

Once you entered the raw scores, the calculator does the rest.
Hope it helps.