The marks of 2376 students in an IB exam are normally distributed with mean 49 marks and variance 225
A) if the mark is 45 estimate the number of candidates who passed the exam?
B) if top 7% of students are awarded a 7, find the minimum mark required to obtain a 7?
mean=49
variance=225
standard deviation=15
n=2376
I don't know how to do this question?
somethung
Anyone find out how to do this?
No problem! I can help you break down the problem and guide you through the process.
A) To estimate the number of candidates who passed the exam with a mark of 45, we need to calculate the z-score and refer to the standard normal distribution table. Here's how you can do it:
1. Calculate the z-score using the formula: z = (x - μ) / σ
- x is the value you want to calculate the z-score for (in this case, 45).
- μ is the mean of the distribution (49 in this case).
- σ is the standard deviation of the distribution (15 in this case).
So, z = (45 - 49) / 15 = -0.27
2. Look up the z-score in the standard normal distribution table or use a calculator. The table gives you the proportion of values below a certain z-score. In this case, we want the proportion of values above the z-score (-0.27), which is equal to the proportion of students who passed the exam.
3. Refer to the table or use a calculator to find the proportion. Let's say the proportion of values below -0.27 is 0.3944 (corresponding to the area under the normal curve).
4. Subtract the proportion from 1 to find the proportion above the z-score:
Proportion above = 1 - 0.3944 = 0.6056
5. Finally, estimate the number of candidates who passed the exam by multiplying the proportion above by the total number of students (n = 2376):
Number of candidates who passed = 0.6056 * 2376 = 1440 (approximately)
Therefore, the estimated number of candidates who passed the exam with a mark of 45 is approximately 1440.
B) To find the minimum mark required to obtain a 7 (top 7% of students), we need to find the corresponding z-score and then use it to find the actual mark. Here's how you can do it:
1. Determine the z-score that corresponds to the top 7% of the distribution. To do this, either use a standard normal distribution table or a calculator that can compute inverse normal probabilities (also known as the percentile function). In this case, we want the z-score that gives an area of 0.07 to the left.
2. Let's say you find that the z-score corresponding to the top 7% is 1.44 (approximately).
3. Calculate the actual mark using the z-score formula:
x = μ + (z * σ)
- μ is the mean of the distribution (49 in this case).
- σ is the standard deviation of the distribution (15 in this case).
- z is the z-score you found in step 2.
So, x = 49 + (1.44 * 15) = 69.6 (approximately)
Therefore, the minimum mark required to obtain a 7 (top 7% of students) is approximately 69.6.
I hope this helps! Let me know if you have any further questions.