If V is the variance and M is the mean of first 15 natural numbers, then what is V + M 2equal to?
Understanding Variance and Mean of Natural Numbers
This question asks us to calculate the sum of the variance and the square of the mean for the first 15 natural numbers. Natural numbers start from 1, so the first 15 natural numbers are 1, 2, 3, ..., up to 15. We need to find the mean (M) and the variance (V) of this set of numbers and then compute the value of \(V + M^2\).
Calculating the Mean (M) of the First 15 Natural Numbers
The mean of the first \(n\) natural numbers is the sum of these numbers divided by \(n\). The sum of the first \(n\) natural numbers is given by the formula \(\frac{n(n+1)}{2}\).
So, the mean M is: \(M = \frac{\text{Sum of first } n \text{ natural numbers}}{n}\) \(M = \frac{\frac{n(n+1)}{2}}{n}\) \(M = \frac{n(n+1)}{2n}\) \(M = \frac{n+1}{2}\)
For the first 15 natural numbers, \(n = 15\). \(M = \frac{15 + 1}{2} = \frac{16}{2} = 8\) So, the mean of the first 15 natural numbers is 8.
Calculating the Variance (V) of the First 15 Natural Numbers
The variance of a set of numbers measures how spread out the numbers are from the mean. For the first \(n\) natural numbers, the variance (V) can be calculated using the formula: \(V = \frac{n^2 - 1}{12}\)
For the first 15 natural numbers, \(n = 15\). \(V = \frac{15^2 - 1}{12}\) \(V = \frac{225 - 1}{12}\) \(V = \frac{224}{12}\)
We can simplify the fraction \(\frac{224}{12}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 4. \(V = \frac{224 \div 4}{12 \div 4} = \frac{56}{3}\) So, the variance of the first 15 natural numbers is \(\frac{56}{3}\).
Calculating V + M²
We have found M = 8 and V = \(\frac{56}{3}\). Now we need to calculate \(V + M^2\). First, let's find \(M^2\). \(M^2 = 8^2 = 64\)
Now, add V and \(M^2\): \(V + M^2 = \frac{56}{3} + 64\)
To add these values, we need to express 64 as a fraction with a denominator of 3. \(64 = \frac{64 \times 3}{3} = \frac{192}{3}\)
Now perform the addition: \(V + M^2 = \frac{56}{3} + \frac{192}{3} = \frac{56 + 192}{3}\) \(V + M^2 = \frac{248}{3}\)
Thus, the value of \(V + M^2\) for the first 15 natural numbers is \(\frac{248}{3}\).
Summary of Calculations
| Quantity | Formula (for first \(n\) natural numbers) | Value for \(n=15\) |
|---|---|---|
| Mean (M) | \(\frac{n+1}{2}\) | \(\frac{15+1}{2} = 8\) |
| Variance (V) | \(\frac{n^2-1}{12}\) | \(\frac{15^2-1}{12} = \frac{224}{12} = \frac{56}{3}\) |
| \(M^2\) | N/A | \(8^2 = 64\) |
| \(V + M^2\) | N/A | \(\frac{56}{3} + 64 = \frac{56}{3} + \frac{192}{3} = \frac{248}{3}\) |
Revision Table: Key Formulas for Natural Numbers
| Measure | Formula for first \(n\) natural numbers |
|---|---|
| Sum | \(\frac{n(n+1)}{2}\) |
| Mean | \(\frac{n+1}{2}\) |
| Variance | \(\frac{n^2-1}{12}\) |
Additional Information on Mean and Variance
The mean is a measure of central tendency, representing the average value in a dataset. The variance is a measure of dispersion, indicating how spread out the data points are around the mean. A higher variance means the data points are more spread out, while a lower variance means they are closer to the mean. Standard deviation, which is the square root of the variance, is also commonly used as a measure of dispersion because it is in the same units as the data. These concepts are fundamental in statistics and probability.
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