Problem 47 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the mean of the data \(: 7,8,9,7,8,7, \lambda, 8\) is 8, then the variance of this data is \(\quad\) [Online April 15, 2018] (a) \(\frac{9}{8}\) (b) 2 (c) \(\frac{7}{8}\) (d) 1

Step-by-Step Solution

Answer

The variance of the data is 1.

Step 1: Calculate the Mean

The mean of given numbers is 8. Let's denote the missing number by \( \lambda \). The formula to compute the mean is \( \frac{\text{sum of all values}}{\text{number of values}} = 8 \). We have 8 terms, so the equation becomes: \[ \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8 \]

Step 2: Solve for \( \lambda \)

Firstly, calculate the sum of all known numbers: \(7 + 8 + 9 + 7 + 8 + 7 + 8 = 54 \). Substitute this into the mean equation to solve for \( \lambda \): \[ \frac{54 + \lambda}{8} = 8 \] Multiply both sides by 8: \[ 54 + \lambda = 64 \]. Then solving for \( \lambda \) gives: \( \lambda = 64 - 54 = 10 \).

Step 3: Calculate Squared Deviations

Now that we have all data points \( 7, 8, 9, 7, 8, 7, 10, 8 \), calculate the deviation for each from the mean and square them: \((7-8)^2, (8-8)^2, (9-8)^2, (7-8)^2, (8-8)^2, (7-8)^2, (10-8)^2, (8-8)^2\). The calculatons are - \((7-8)^2 = 1\) (appears thrice) - \((9-8)^2 = 1\)- \((8-8)^2 = 0\) (appears thrice)- \((10-8)^2 = 4\).

Step 4: Calculate the Variance

Sum the squared deviations: \(1 + 0 + 1 + 1 + 0 + 1 + 4 + 0 = 8\). Since variance is the mean of these squared deviations, divide by 8: \[ \frac{8}{8} = 1 \]. The variance is therefore \( 1 \).

Key Concepts

Mean

To find the mean, or average, of a set of numbers, you add up all the numbers and then divide by the total number of items in the set. It's a way to figure out the central value of the data. Imagine if we had a data set with the numbers: 7, 8, 9, 7, 8, 7, a missing number denoted as \( \lambda \), and 8. The mean of this set is 8.
We use this handy formula for calculating the mean:

• Sum of all the numbers
• Divided by the count of the numbers

This is given as: \[ \frac{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8}{8} = 8 \]
From this equation, we find that when you add all the numbers together and divide by 8, you get the mean, which is a central idea in understanding your data set.

Squared Deviations

Squared deviations help us see how much each data point "deviates" or differs from the mean. After figuring out all our data points, including the missing number \( \lambda \) as 10, we get a complete list: 7, 8, 9, 7, 8, 7, 10, 8. Now, we compute the squared deviation for each number.
Each data point subtraction from the mean (in this case, 8), and then squaring it, tells us how far it lies from the mean. Let's see how this works:

• \((7 - 8)^2 = 1\) (appears three times)
• \((9 - 8)^2 = 1\)
• \((8 - 8)^2 = 0\) (appears three times)
• \((10 - 8)^2 = 4\)

The squaring makes every deviation positive, which simplifies the comparison among these deviations. It allows us to emphasize larger differences by making them even larger.

Variance Calculation

Variance provides a measure of how much the data are spread out around the mean.
To calculate the variance, we first sum all the squared deviations obtained in the previous step: \(1 + 0 + 1 + 1 + 0 + 1 + 4 + 0 = 8\).
Since we have 8 data points, the variance is calculated by finding the average of these squared deviations. This means dividing the total sum of squared deviations by the number of data points: \[ \frac{8}{8} = 1 \]
The variance, therefore, is 1. This tells us the average of how much each number in our data set differs from the mean. It is a useful indicator of the dataset's variability or spread. The greater the variance, the wider the spread, and vice versa.