Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A scientist is weighing each of 30 fishes. Their mean weight worked out is \(30 \mathrm{gm}\) and a standarion deviation of \(2 \mathrm{gm}\). Later, it was found that the measuring scale was misaligned and always under reported every fish weight by \(2 \mathrm{gm}\). The correct mean and standard deviation (in gm) of fishes are respectively: (a) 32,2 (b) 32,4 (c) 28,2 (d) 28,4

Step-by-Step Solution

Answer

The correct mean is 32 gm and the standard deviation remains 2 gm (Option a).

Step 1: Identify the Problem

The problem indicates that each fish's weight was underreported by 2 grams. We need to adjust the mean and understand how this shift affects the standard deviation.

Step 2: Adjust the Mean Weight

The initial mean weight was 30 gm, but since each fish was underreported by 2 gm, to get the correct mean, we add 2 gm to this mean.\[ \text{Correct mean} = 30 + 2 = 32 \text{ gm} \]

Step 3: Understand the Impact on Standard Deviation

The standard deviation measures the spread of the weights around the mean. A constant underreporting (or overreporting) of each value does not affect the spread or variability. Therefore, the standard deviation remains the same.

Step 4: Find the Correct Standard Deviation

Since the standard deviation is unaffected by adding or subtracting a constant to each data point, the standard deviation remains 2 gm.

Step 5: Compile the Correct Results

The corrected mean is 32 gm, and the standard deviation remains at 2 gm. Thus, the correct answer is option (a): 32, 2.

Key Concepts

Mean

The mean, often referred to as the average, is a fundamental concept in statistics. It gives a central value of a dataset and is calculated by summing up all values and then dividing by the number of values. In this example, the mean weight of fish was initially computed as 30 grams.
The interesting aspect of mean calculation is its sensitivity to shifts in data. If each data point in a dataset is increased by a constant amount, the mean will also increase by that same amount. Therefore, when the scale misalignment was discovered, adding the 2 grams back to each fish's weight led to the new mean:
\[ \text{Correct mean} = 30 + 2 = 32 \text{ gm} \]
Understanding how the mean is affected by changes in data is essential for accurately interpreting and adjusting data. This is particularly important in scientific experiments where precision matters.

Standard Deviation

Standard deviation is a measure of the spread or dispersion within a set of data. It tells us how much the individual data points tend to deviate from the mean. In simpler terms, it gives us insights into the variation around the average value. For our fish weight example, the initial standard deviation was 2 grams.
A unique feature of standard deviation is that it remains unchanged by the addition or subtraction of a constant from each data point in a dataset. This is because such an operation doesn’t alter the relative distances between data points. So, even if every fish's weight was increased by 2 grams to correct the scale error, the standard deviation stays constant at 2 grams.
This property makes standard deviation a reliable measure of variability, unaffected by shifts in data caused by erroneous measurements.

Data Correction

Data correction is the process of identifying and rectifying errors within a dataset. Performing data correction is crucial to ensure the accuracy and reliability of results derived from the data.
In the given exercise, data correction involved adjusting the reported weights of fish by adding 2 grams to each measurement. This correction impacted the calculated mean but left the standard deviation unaffected, given no real change in variability was introduced by the consistent offset.

• Identify the error: The scale was misaligned by 2 grams.
• Adjust the recorded data accordingly: Add the missing 2 grams to the mean.
• Review other statistical measures: Recognize that standard deviation remains unchanged in such corrections.

This example underscores the importance of regular equipment calibration and verification in scientific measurements. Being vigilant about potential data errors and understanding their correction ensures credibility and precision in statistical analysis.