Problem 38 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A student scores the following marks in five tests: 45,54 , \(41,57,43\). His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is : [April. \(\mathbf{0 8}, \mathbf{2 0 1 9}\) (II)] (a) \(\frac{10}{\sqrt{3}}\) (b) \(\frac{100}{3}\) (c) \(\frac{10}{3}\) (d) \(\frac{100}{\sqrt{3}}\)

Step-by-Step Solution

Answer

The standard deviation of the marks in six tests is \(\frac{10}{\sqrt{3}}\), so the answer is (a).

Step 1: Calculate total marks for six tests

Since the mean score of the six tests is given as 48, first calculate the total marks for all six tests. The mean is calculated by dividing the total by the number of tests, which is 6: \[ \text{Total score} = \text{Mean} \times \text{Number of tests} = 48 \times 6 = 288 \]

Step 2: Calculate the sum of known scores

Add the scores from the five known tests: 45, 54, 41, 57, and 43.\[ 45 + 54 + 41 + 57 + 43 = 240 \]

Step 3: Find the unknown score for the sixth test

Calculate the score in the sixth test by subtracting the sum of the scores of the first five tests from the total score needed for six tests to maintain the mean of 48.\[ \text{Score of sixth test} = 288 - 240 = 48 \]

Step 4: Calculate variance

The variance is the average of the squared differences from the mean. Calculate the variance for the scores 45, 54, 41, 57, 43, and 48. First, find each score's difference from the mean (48), square it, and then average these squared differences.The individual differences squared are:- For 45: \[ (45 - 48)^2 = 9 \]- For 54: \[ (54 - 48)^2 = 36 \]- For 41: \[ (41 - 48)^2 = 49 \]- For 57: \[ (57 - 48)^2 = 81 \]- For 43: \[ (43 - 48)^2 = 25 \]- For 48: \[ (48 - 48)^2 = 0 \]Calculate the average of these squared differences:\[ \frac{9 + 36 + 49 + 81 + 25 + 0}{6} = \frac{200}{6} \]

Step 5: Calculate standard deviation

The standard deviation is the square root of the variance. Find the standard deviation by taking the square root of the variance obtained.\[ \sqrt{\frac{200}{6}} = \frac{10}{\sqrt{3}} \]

Step 6: Select the closest answer

The calculated standard deviation is \(\frac{10}{\sqrt{3}}\), so the correct answer choice from the given options is (a) \(\frac{10}{\sqrt{3}}\).

Key Concepts

Mean Score

The mean score is an important concept in statistics. It is the average of a set of numbers and gives a single value that represents the entire dataset. Let's explain how it works using the example from the exercise.
The student has scores from five tests and requires a sixth test to achieve an average (mean) score of 48. The formula to calculate the mean score is:

• Mean = Total Score / Number of Tests

A key step is calculating the total score needed for all six tests to achieve this mean. As we know:

• Total Score = Mean x Number of Tests = 48 x 6 = 288

Thus, the total marks must be 288 for an average of 48 across six tests.
This gives a useful perspective on how additional data points influence the mean in statistics.

Variance

Variance is a statistical measure that expresses how much the values in a dataset differ from the mean. It helps in quantifying the degree of variation or dispersion in a dataset. Here’s how variance is determined in the exercise:

• First, find the difference between each score and the mean.
• Next, square these differences to eliminate negative values.
• Finally, average the squared differences to find the variance.

For the student’s scores, subtract each score from the mean (48), square each result, and compute the mean of these squared differences. Using the exercise's data:

• Variance = (9 + 36 + 49 + 81 + 25 + 0) / 6 = 200 / 6

This calculation gives insight into how much variation exists among the test scores. A higher variance signifies wide dispersion, while a lower variance indicates that the data points are more similar to each other.

Mathematics Problems

Solving mathematics problems often involves applying various statistical methods. In this case, students must determine the missing data point to ensure a specific mean score. Here's a breakdown of the problem-solving strategy:

• Understand the given and what needs to be found: known scores, desired mean, and unknown test score.
• Use mathematical formulas: calculate total required score and deduce unknown score.
• Apply statistical formulas: determine variance and standard deviation.

This approach not only involves numerical calculations but also helps students develop critical thinking skills. Solving such math problems enhances one's ability not only to work with numbers but also to develop a systematic approach to problem-solving.

Statistics Concepts

Statistics concepts such as mean, variance, and standard deviation offer a structured way of understanding data variability. They are interconnected and often used together to provide a comprehensive picture of data behavior.
1. **Mean**: Central tendency indicating an average. 2. **Variance**: Measure of dispersion highlighting how scores vary from the mean.
3. **Standard Deviation**: Indicates the spread of data points around the mean, derived from variance by taking its square root.

• This particular exercise asks for the standard deviation, guiding students through variance calculation first.
• Standard Deviation = √Variance = √(200/6) = 10/√3

These statistics concepts are key to analyzing and interpreting data, helping students make informed decisions and predictions based on data patterns.