Step 1 :Given that the mean reading speed is 88 wpm and the standard deviation is 10 wpm, we are asked to find the probability that a random sample of 24 second grade students from the city results in a mean reading rate of more than 92 words per minute.
Step 2 :This is a problem of normal distribution. We can use the z-score formula to calculate the z-score for 92 wpm, and then use the standard normal distribution table to find the probability.
Step 3 :The z-score formula is: \(z = \frac{X - \mu}{\sigma / \sqrt{n}}\), where \(X\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size. In this case, \(X = 92\), \(\mu = 88\), \(\sigma = 10\), and \(n = 24\).
Step 4 :After calculating the z-score, we can find the probability that a random sample of 24 students has a mean reading rate of more than 92 wpm by looking up the z-score in the standard normal distribution table. However, the table gives us the probability that the sample mean is less than 92 wpm, so we need to subtract this probability from 1 to get the probability that the sample mean is more than 92 wpm.
Step 5 :Let's calculate the z-score and find the probability.
Step 6 :Using the given values in the z-score formula, we get \(z = 1.96\).
Step 7 :Looking up the z-score in the standard normal distribution table, we get a probability of 0.0250.
Step 8 :Therefore, the probability that a random sample of 24 second grade students from the city results in a mean reading rate of more than 92 words per minute is approximately 0.0250.
Step 9 :Final Answer: The probability is \(\boxed{0.0250}\).