Final answer:
The question involves using the Central Limit Theorem and probabilities to find how many students must take an exam so that with a 70% probability the class average is between 50 and 70. It requires calculating the sample size needed to ensure that the class average deviation is within one standard deviation of the mean for a normal distribution.
Step-by-step explanation:
The subject of the question is the determination of how many students need to take a final exam to ensure that the class average falls within a certain range with a given probability. Specifically, it involves using the Central Limit Theorem and knowledge of probabilities to find the sample size required. In this scenario, we are given the mean score and the variance of the final exam scores from prior experience and asked to calculate the sample size (number of students) needed to ensure with a probability of 0.7 (70%) that the class average would be within the interval [50, 70].
To apply the Central Limit Theorem, we use the mean score (μ = 65) and the variance (σ² = 12) to find the standard deviation (σ). First, we take the square root of the variance, which gives us the standard deviation of the scores. Then, to find the standard deviation of the sampling distribution of the class average, we divide the standard deviation of the scores by the square root of the sample size (n). We use the fact that for a normally distributed variable, the probability of a value lying within one standard deviation of the mean is about 68%, within two standard deviations is about 95%, and within three standard deviations is about 99.7%. In this case, we are aiming for a 70% probability, so we want our calculated interval to be within roughly one standard deviation of the sampling distribution's mean.
We can express this mathematically by setting up a proportion that allows us to solve for the sample size. This involves equating the desired probability corresponding to the z-score in a standard normal distribution to our scenario. Being within [50, 70] means that the class average could deviate from the mean by 15 points on either side, which needs to be less than or equal to one standard deviation of the sample's mean distribution for a 70% probability.
Without reaching for specifics on calculations involving z-scores, it's clear that advanced knowledge in statistics is required to tackle this problem step by step. The exact solution would involve using the standard normal distribution, z-tables, and the desired probability to find the z-score and then solve for the sample size 'n' that makes the class average fall within the required confidence interval.