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# 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 k^{th} 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 k^{th} 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 | 4k^{2} | 6k^{2} | 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 + 4k^{2} + 6k^{2} + k = 1 (Putting the P(x)-values from the chart)

⇒ 2k + 10k^{2} + k = 1

⇒ 3k + 10k^{2} = 1

⇒ 3k + 10k^{2} – 1 = 0

⇒ 10k^{2} + 3k – 1 = 0

⇒ 10k^{2} + 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).