Problem 48 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\sum_{\mathrm{i}=1}^{9}\left(\mathrm{x}_{\mathrm{i}}-5\right)=9\) and \(\sum_{\mathrm{i}=1}^{9}\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}=45\), then the standard deviation of the 9 items \(x_{1}, x_{2}, \ldots, x_{9}\) is: [2018] (a) 4 (b) 2 (c) 3 (d) 9

Step-by-Step Solution

Answer

The standard deviation is approximately 2.236, closest to option (b) 2.

Step 1: Understanding the Given Information

We are given two important equations to deal with the set of numbers \( \{x_1, x_2, \ldots, x_9\} \). First, \( \sum_{i=1}^{9}(x_i - 5) = 9 \), and second, \( \sum_{i=1}^{9}(x_i - 5)^2 = 45 \). These equations provide insight into the means and variance calculations of these numbers.

Step 2: Relate the Mean to Given Information

The equation \( \sum_{i=1}^{9}(x_i - 5) = 9 \) can be rewritten as \( \sum_{i=1}^{9}x_i - 5 \times 9 = 9 \). This simplifies to \( \sum_{i=1}^{9}x_i = 54 \). Thus, the mean of the 9 items is \( \frac{54}{9} = 6 \).

Step 3: Calculate Variance Using Provided Equations

The equation \( \sum_{i=1}^{9}(x_i - 5)^2 = 45 \) is directly related to the variance of the items shifted by 5. We already know that the mean of the items is 6, thus, the variance is \( \frac{1}{9} \sum_{i=1}^{9}(x_i - 6)^2 = \frac{45}{9} = 5 \).

Step 4: Determine the Standard Deviation

The standard deviation is the square root of the variance we calculated. Therefore, the standard deviation is \( \sqrt{5} \approx 2.236 \). Since the closest standard deviation in the options is \(2\), the correct choice is (b) 2.

Key Concepts

Variance

Variance is a measure of how much the values in a data set differ from the mean. It gives us insight into the spread or dispersion of the numbers within the set. In our exercise, we used the equation \( \sum_{i=1}^{9}(x_i - 5)^2 = 45 \) to calculate the variance.
This equation illustrates how much each number, shifted by 5, deviates squared from that shift. We already calculated that the variance is \( \frac{45}{9} = 5 \) when relating the data back to its actual mean of 6. This variance indicates that, on average, each number lies roughly \( \sqrt{5} \) units away from the mean.
This calculation is crucial because it represents the degree to which numbers in your dataset vary compared to the mean, thus providing a foundation for the standard deviation measurement.
Understanding variance is key to interpreting the spread of your data and predicting future events accurately, as a higher variance would mean more scattered data, while a lower variance suggests that the data points are closer to the mean.

Mean

The mean, often referred to as the average, is the sum of all the numbers in a data set divided by the number of numbers. It's a central value representing your data set. In our exercise, we found the mean by summing all nine data points \( \sum_{i=1}^{9}x_i = 54 \), and then dividing by 9.
This gave us the mean: \( \frac{54}{9} = 6 \).
The mean is useful because it provides a simple summary statistic that has various applications, such as assessing central trends or comparing different datasets. While the mean is easy to calculate, it can be misleading in the presence of outliers, which are extreme values that can skew the average. Therefore, it's also essential to consider other statistics like the variance or median when describing a dataset.

• It reflects a central value where data points are balanced.
• Sensitive to extreme values, hence best used with other metrics like variance.

Sum of Squares

The Sum of Squares is a crucial step in calculating both variance and standard deviation. It helps in understanding the spread of your data by quantifying the total deviation from a chosen mean. In our exercise, this was represented by the equation \( \sum_{i=1}^{9}(x_i - 5)^2 = 45 \).
Here, each data point was adjusted by subtracting a given number (5 in this case), followed by squaring the result. This squaring is crucial because it prevents negative and positive deviations from canceling each other out. Overall, the Sum of Squares helps determine how much variability exists that is unexplained by the mean alone.

• Preliminary step in variance calculation to avoid cancellation of deviations.
• Captures the extent of deviation each data point has from the calculated mean.

Understanding Sum of Squares is crucial for deeper statistical analysis, such as in ANOVA or linear regression models, as it aids in the assessment of overall data spread or variance.