Suppose the top 2% of JEE aspirants(students who appeared for JEE advanced exam) will get admission in IIT. If the score of JEE-advanced is normally distributed with mean 500 and standard deviation 100 then how much minimum marks should be score to get admission in IIT.

Group of answer choices

None of these

725

705

750

To find the minimum marks required to be in the top 2% of JEE aspirants, we need to determine the score corresponding to the 98th percentile of a normal distribution with a mean of 500 and a standard deviation of 100.

The steps to solve this problem are as follows:

1. **Determine the z-score corresponding to the top 2%**:
   - The top 2% of the distribution corresponds to the 98th percentile.
   - From standard normal distribution tables, or using a calculator, the z-score corresponding to the 98th percentile is approximately \( z = 2.05 \).

2. **Use the z-score formula to find the corresponding score**:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   where:
   - \( X \) is the score we need to find,
   - \( \mu = 500 \) is the mean,
   - \( \sigma = 100 \) is the standard deviation,
   - \( z = 2.05 \) is the z-score for the 98th percentile.

   Rearrange the formula to solve for \( X \):
   \[
   X = z \cdot \sigma + \mu
   \]
   Substitute the values:
   \[
   X = 2.05 \times 100 + 500 = 205 + 500 = 705
   \]

Therefore, the minimum marks required to be in the top 2% of JEE aspirants, and thus get admission to IIT, is **705**.

So the correct answer is \( \boxed{705} \).

Would you like more details or have any questions? Here are some related questions:

1. How can the z-score be interpreted in a normal distribution?
2. What is the significance of the standard deviation in determining scores?
3. How would the required score change if the standard deviation was different?
4. What percentile corresponds to being in the top 1% of the distribution?
5. How can one compute probabilities for non-normal distributions?

**Tip:** Always double-check the z-score corresponding to specific percentiles when working with normal distributions, as even a small error in the z-score can lead to a significant difference in the calculated result.

Learn how to calculate the minimum marks required to be in the top 2% of JEE aspirants and secure admission to IIT. Understand the concept of normal distribution, percentiles, and z-scores with a step-by-step solution. Suitable for high school students preparing for competitive exams.

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