If the standard deviation of the numbers `-1, 0,1,k` is `sqrt(5)` where `k gt 0` , then k is equal to `2 sqrt(l)` . The value of `l` is ________.

Explanation

To solve the problem, we need to find the value of \( k \) given that the standard deviation of the numbers \(-1, 0, 1, k\) is \(\sqrt{5}\). We will follow these steps:

### Step 1: Understand the formula for standard deviation
The standard deviation \( \sigma \) of a set of numbers is given by the formula:
\[ \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}} \]
where \( \mu \) is the mean of the numbers, and \( n \) is the number of terms.

### Step 2: Calculate the mean (\( \mu \))
The numbers we have are \(-1, 0, 1, k\). The mean \( \mu \) is calculated as follows:
\[ \mu = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4} \]

### Step 3: Calculate the sum of squares
Next, we need to calculate the sum of squares of the deviations from the mean:
\[ \sum (x_i - \mu)^2 = (-1 - \mu)^2 + (0 - \mu)^2 + (1 - \mu)^2 + (k - \mu)^2 \]
Substituting \( \mu = \frac{k}{4} \):
\[ = \left(-1 - \frac{k}{4}\right)^2 + \left(0 - \frac{k}{4}\right)^2 + \left(1 - \frac{k}{4}\right)^2 + \left(k - \frac{k}{4}\right)^2 \]

### Step 4: Expand the squares
Now we expand each term:
1. \(\left(-1 - \frac{k}{4}\right)^2 = \left(-\frac{4 + k}{4}\right)^2 = \frac{(4 + k)^2}{16}\)
2. \(\left(0 - \frac{k}{4}\right)^2 = \left(-\frac{k}{4}\right)^2 = \frac{k^2}{16}\)
3. \(\left(1 - \frac{k}{4}\right)^2 = \left(\frac{4 - k}{4}\right)^2 = \frac{(4 - k)^2}{16}\)
4. \(\left(k - \frac{k}{4}\right)^2 = \left(\frac{3k}{4}\right)^2 = \frac{9k^2}{16}\)

### Step 5: Combine the terms
Now we can combine these:
\[ \sum (x_i - \mu)^2 = \frac{(4 + k)^2 + k^2 + (4 - k)^2 + 9k^2}{16} \]
Expanding each term:
- \((4 + k)^2 = 16 + 8k + k^2\)
- \((4 - k)^2 = 16 - 8k + k^2\)

Combining these gives:
\[ = \frac{(16 + 8k + k^2) + k^2 + (16 - 8k + k^2) + 9k^2}{16} = \frac{32 + 3k^2 + 9k^2}{16} = \frac{32 + 11k^2}{16} \]

### Step 6: Set up the equation for standard deviation
Now we set the equation for standard deviation:
\[ \sqrt{5} = \sqrt{\frac{32 + 11k^2}{16 \cdot 4}} = \sqrt{\frac{32 + 11k^2}{64}} \]
Squaring both sides:
\[ 5 = \frac{32 + 11k^2}{64} \]
Multiplying both sides by 64:
\[ 320 = 32 + 11k^2 \]
Subtracting 32 from both sides:
\[ 288 = 11k^2 \]
Dividing by 11:
\[ k^2 = \frac{288}{11} \]

### Step 7: Find \( k \)
Taking the square root:
\[ k = \sqrt{\frac{288}{11}} = \frac{\sqrt{288}}{\sqrt{11}} = \frac{12\sqrt{2}}{\sqrt{11}} = 2\sqrt{6} \]

### Step 8: Identify \( l \)
Given that \( k = 2\sqrt{l} \), we can compare:
\[ 2\sqrt{l} = 2\sqrt{6} \implies \sqrt{l} = \sqrt{6} \implies l = 6 \]

### Final Answer
The value of \( l \) is \( \boxed{6} \).