Step 1 :The problem is asking for the probability that the mean of a sample would differ from the population mean by less than 2.5 points. This is a question about the sampling distribution of the mean.
Step 2 :The Central Limit Theorem tells us that the sampling distribution of the mean is normally distributed with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.
Step 3 :We can calculate the z-score for a difference of 2.5 points from the mean using the formula \(z = \frac{x - \mu}{\sigma / \sqrt{n}}\), where \(x\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
Step 4 :The z-score tells us how many standard deviations away from the mean our value is. We can then use a z-table or a function like scipy's norm.cdf to find the probability that a z-score is less than our calculated value.
Step 5 :Since the question asks for the probability that the mean differs by less than 2.5 points, we need to find the probability for both 2.5 points above and below the mean and add them together.
Step 6 :Let's calculate this: \(\mu = 187\), \(\sigma = 13\), \(n = 62\), \(x1 = 184.5\), \(x2 = 189.5\), \(z1 = -1.5142322834638098\), \(z2 = 1.5142322834638098\), \(p1 = 0.06498346820877393\), \(p2 = 0.935016531791226\), \(p = 0.870033063582452\)
Step 7 :The calculated probability that the mean of the sample would differ from the population mean by less than 2.5 points is approximately 0.8700.
Step 8 :Final Answer: The probability that the mean of the sample would differ from the population mean by less than 2.5 points if 62 exams are sampled is approximately \(\boxed{0.8700}\).