Contents
Meaning of k in statistics
The k value has a very different definition in different modules of statistics and mathematics in general. We will try to identify the 3 major k values in statistics here:
Defining k in the sense of computing combinations
Sample problem to solve:
Let’s say there are 5 different color marbles. Blue, Red, Green, Yellow, and Purple.
Now if you randomly choose 3 different marbles from these 5, how many different combinations there can be?
Answer:
Here, n = 5 (i.e. the total number of marbles)
k = 3 (i.e. the number of marbles you will have to randomly choose).
(here, 0 ≦ k ≦ n)
The number of different combinations there can be found by the equation:
| C (n, k) = n! / {(n-k)! k!} |
⇒ C (5,3) = 5! / {(5-3)! 3!} (putting the values of n and k)
⇒ C (5,3) = 5! / (2! 3!)
⇒ C (5,3) = (5x4x3x2x1) / {(2×1) (3x2x1)}
⇒ C (5,3) = 120/12
⇒ C (5,3) = 10
So the answer is there can be 10 different combinations, when you choose 3 (i.e. the k value) from a total of 5 different color marbles. (This is the answer).
Defining k in the sense of Systematic Sampling
Sample problem to solve (1):
Let’s say there are 500 students on an MBA program at a university. Now you have to make a systematic sample consisting of 10 students with every kth student on the student list, after randomly selecting a starting point from 1 to k.
Answer:
Here, n = 500 (i.e the total number of students)
No. of units (i.e. students) in the sample = 10
To find the value of k:
| k = n / (no. of units in the sample) |
⇒ k= 500/10
⇒ k= 50
Now, we will have to select a starting point from 1 to k, i.e. from 1 to 50.
Let’s say we have selected no. 3.
So, the sample would be:
3, (3+k), ((3+k)+k), (((3+k)+k)+k),….…… till (3+9k).
⇒ 3, (3+50), ((3+50)+50), (((3+50)+50)+50),……… till {3+(950)}.
⇒ 3, 53, 103, 153, 203, 253, 303, 353, 403, 453.
(This is the answer).
Sample problem to solve (2):
Let’s say there are 50 houses on the street of your neighbourhood. Now, you have to make a systematic sample consisting of 9 houses with every kth house on the street, after randomly selecting a starting point from 1 to k.
Answer:
Here, n = 50 (i.e the total number of houses)
No. of units (i.e. houses) in the sample = 9
To find the value of k
| k = n / (no. of units in the sample) |
⇒ k= 50/9
⇒ k= 5.56
⇒ k= 6 (Rounding the value. Rounding may occur changes like 1 in the sample size than the correct/desired value).
Now, we will have to select a starting point from 1 to k, i.e. from 1 to 6.
Let’s say we have selected no. 5.
So, the sample would be:
5, (5+k), ((5+k)+k), (((5+k)+k)+k),….…… till (5+8k).
⇒ 5, (5+6), ((5+6)+6), (((5+6)+6)+6),……… till {5+(86)}.
⇒ 5, 11, 17, 23, 29, 35, 41, 47, 53. (It is more than the total no. of houses i.e. 50, as we rounded the value of k).
(This is the answer).
Defining k in the Probability distribution
Sample problem to solve (1):
The random variable x has the probability distribution as below. Determine the value of k.
| x | 0 | 1 | 2 | 3 |
| P(x) | k | 3k | 4k | 9k |
Answer:
We need to satisfy 2 conditions here:
CONDITION 1: Each probability of x has to have a value between 0 and 1.
| 0 ≦ P(x) ≦ 1 |
i.e. we can’t have negative numbers here or numbers more than 1.
CONDITION 2: The sum of probabilities must be equal to 1.
| ∑P(x) = 1 |
Now from the second condition:
∑P(x) = 1
⇒ P(0) + P(1) + P(2) + P(3) = 1 (Putting the x-values from the chart)
⇒ k + 3k + 4k + 9k = 1 (Putting the P(x)-values from the chart)
⇒ 17k = 1
⇒ k = 1/17
(This is the answer).
Sample problem to solve (2):
Now in this sample, we are going to see how condition 1 explained in the above example applies while determining the k value (i.e. the condition 0 ≦ P(x) ≦ 1). So, this example is going to be a bit more complex than the above one.
The random variable x has the probability distribution as below. Determine the value of k.
| x | -1 | 0 | 1 | 2 |
| P(x) | 2k | 4k2 | 6k2 | k |
Answer:
Now from the second condition:
∑P(x) = 1
⇒ P(-1) + P(0) + P(1) + P(2) = 1 (Putting the x-values from the chart)
⇒ 2k + 4k2 + 6k2 + k = 1 (Putting the P(x)-values from the chart)
⇒ 2k + 10k2 + k = 1
⇒ 3k + 10k2 = 1
⇒ 3k + 10k2 – 1 = 0
⇒ 10k2 + 3k – 1 = 0
⇒ 10k2 + 5k – 2k – 1 = 0
⇒ 5k (2k + 1) – 1 (2k + 1) = 0
⇒ (5k – 1) (2k + 1) = 0
Therefore,
(5k – 1) = 0 and (2k + 1) = 0
⇒ k = 1/5 and 2k = -1
⇒ k =1/5 and k = – 1/2
Applying Condition 1: 0 ≦ P(x) ≦ 1:
⇒ k = 1/5 (✅Valid) and k = – 1/2 (❌Not valid; As the probability of x cannot be a negative number).
Therefore, k = 1/5
(This is the answer).