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Chapter 23: Problem 46

Let \(x_{1}, x 2, \ldots, x_{n}\) be \(n\) observations such that \(\sum x_{i}^{2}=400\) and \(\sum x_{i}=80 .\) Then a possible value of \(n\) among the following is [2005] (A) 15 (B) 18 (C) 9 (D) 12

Short Answer

Expert verified
The possible value of \( n \) is 18.

Step by step solution

01

Understanding the Given Equations

We are given two equations involving the observations: \( \sum x_i^2 = 400 \) and \( \sum x_i = 80 \). Our task is to find a possible value of \( n \) such that these conditions are satisfied.
02

Using the Formula for Variance

Recall the formula for the variance of a set of numbers: \( \sigma^2 = \frac{1}{n}\sum x_i^2 - \left(\frac{1}{n}\sum x_i\right)^2 \). Substitute the given sums into this formula.
03

Simplifying the Variance Formula

Plug the given values into the variance formula: \( \sigma^2 = \frac{1}{n}(400) - \left(\frac{1}{n}(80)\right)^2 \) which simplifies to: \( \sigma^2 = \frac{400}{n} - \frac{6400}{n^2} \).
04

Checking Positive Variance

The variance must be non-negative, so \( \frac{400}{n} - \frac{6400}{n^2} \geq 0 \). This simplifies to \( 400n \geq 6400 \) leading to \( n \geq 16 \).
05

Finding the Valid Option

From the choices given, the only option that satisfies \( n \geq 16 \) is (B) 18. Thus, 18 is the valid option.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Squares
The sum of squares is a fundamental concept in statistics that is used to measure the total variation in a data set. It is always a non-negative value, indicating the squared differences from the mean or the individual data points. In the context of this exercise, the sum of squares \( \sum x_i^2 = 400 \). This value is calculated by squaring each observation and then adding them all together. Consider it like this:
  • Each observation \( x_i \) is squared: \( x_i^2 \).
  • The squared values are summed up: \( \sum x_i^2 \).
The sum of squares is used in calculating variance and other statistical measures, giving us a sense of the magnitude of the variations or changes in the data set from a specific point, often the mean.
Sum of Observations
The sum of observations refers to the total of all individual data points in a sample or a population. In our problem, this is given as \( \sum x_i = 80 \). Here's what this means:
  • Each observation \( x_i \) directly contributes to the total.
  • The sum is simply the addition of each data point.
Think of it as laying out all the data points on a line and then squishing them together to see how much there is altogether. It's a way to see the size of the data or the overall level of these observations. This total is key when calculating other statistical measures like the mean or variance because it gives a way to account for the collective impact of all data points.
Variance Formula
The variance formula is used to find the variance \( \sigma^2 \) of a set of observations. It gives an idea of how the data is spread out or how much it varies. For a dataset, the formula is given by:\[ \sigma^2 = \frac{1}{n}\sum x_i^2 - \left(\frac{1}{n}\sum x_i\right)^2 \]Breaking it down:
  • \( \frac{1}{n}\sum x_i^2 \) is the average of the squares of the observations.
  • \( \left(\frac{1}{n}\sum x_i\right)^2 \) is the square of the average of the observations.
  • Subtracting these gives a measure of the spread of the data.
In our exercise, substituting the given sums into the variance formula allows us to solve for \( n \), while ensuring a non-negative result is crucial for finding a valid variance.
Non-negative Variance
Variance is always non-negative because it is calculated based on squared terms, which are inherently non-negative. In the exercise above, ensuring that the expression remains non-negative helps confirm that the number of observations \( n \) is valid.For our variance equation:\[ \frac{400}{n} - \frac{6400}{n^2} \geq 0 \]This inequality safeguards that the variance is not less than zero. It can also be interpreted as follows:
  • A non-negative variance indicates that the data points do not deviate from each other by a negative amount.
  • It ensures logical consistency within statistical analysis.
In this problem, solving for \( n \) gives us a minimum threshold 鈥� essentially, how few observations could collectively provide the given sums while maintaining a realistic variance.

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Most popular questions from this chapter

Let \(x_{1}, x_{2} \ldots x_{n}\) be \(n\) observations, and let \(\bar{x}\) be their arithmetic mean and \(\sigma^{2}\) be their variance. Statement-1: Variance of \(2 x_{1}, 2 x_{2} \ldots 2 x_{n}\) is \(4 \sigma^{2}\). Statement-2: Arithmetic mean of \(2 x_{1}, 2 x_{2} \ldots 2 x_{n}\) is 4 \(\bar{x} .\) (A) Statement-1 is false, statement- 2 is true (B) Statement- 1 is true, statement- 2 is true; statement 2 is a correct explanation for statement 1 (C) Statement- 1 is true, statement-2 is true; statement 2 is not a correct explanation for statement 1 (D) Statement-1 is true, statement- 2 is false

If the mean deviation about the median of the numbers \(a, 2 a, \ldots, 50 a\) is 50 , then \(|a|\) equals (A) 5 (B) 2 (C) 3 (D) 4

The mean weight per student in a group of seven students is \(55 \mathrm{~kg}\) If the individual weights of 6 students are \(52,58,55,53,56\) and 54 ; then weight of the seventh student is (A) \(55 \mathrm{~kg}\) (B) \(60 \mathrm{~kg}\) (C) \(57 \mathrm{~kg}\) (D) \(50 \mathrm{~kg}\)

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For three numbers \(a, b, c\) product of the average of the numbers \(a^{2}, b^{2}, c^{2}\) and \(\frac{1}{a^{2}}, \frac{1}{b^{2}}, \frac{1}{c^{2}}\) cannot be less than (A) 1 (B) 3 (C) 9 (D) None of these
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