To solve this, we can use the z-score formula, which helps in converting a given value from a normal distribution into the standard normal distribution. The problem assumes the scores follow a normal distribution with a mean of 120 and a standard deviation of 15.

Step 1: Compute the z-score

The z-score formula is:

$z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the score we are interested in (100 in this case),
• $\mu$ is the mean (120),
• $\sigma$ is the standard deviation (15).

$z = \frac{100 - 120}{15} = \frac{-20}{15} = -1.33$

Step 2: Use the z-score to find the probability

A z-score of -1.33 corresponds to the cumulative probability from the standard normal distribution. We can look this up in a z-table, or use a calculator, to find that the cumulative probability for $z = -1.33$ is approximately 0.0918.

This represents the probability of scoring below 100. Since we are interested in the probability of scoring above 100, we subtract this from 1:

$P(X > 100) = 1 - 0.0918 = 0.9082$

Final Answer:

The probability that BIT students scored above 100 is approximately 0.9082, or 90.82%.

Would you like further details or have any questions about this?

Here are five related questions you might find interesting:

1. What is the probability of scoring below 110 in the contest?
2. How do we calculate the probability for scores above 130?
3. If the standard deviation changed, how would it affect the probability of scores above 100?
4. How would you calculate the percentage of students who scored between 105 and 125?
5. Can you explain how the normal distribution applies in contests like these?

Tip: In probability questions involving normal distributions, always convert your raw score to a z-score to use standard normal distribution tables effectively.