The reading speed of second grade students in a large city is approximately normal, with a mean of 92 words per minute (wpm) and a standard deviation of $10 \mathrm{wpm}$. Complete parts (a) through (f). Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Interpret this probability. Select the correct choice below and fill in the answer box within your choice. A. If 100 different samples of $n=20$ students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of less than 98 words per minute. B. If 100 different samples of $n=20$ students were chosen from this population, we would expect 0 sample(s) to have a sample mean reading rate of more than 98 words per minute. C. If 100 different samples of $n=20$ students were chosen from this population, we would expect sample(s) to have a sample mean reading rate of exactly 98 words per minute. (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. A. Increasing the sample size decreases the probability because $\sigma_{\bar{x}}$ decreases as $n$ increases. B. Increasing the sample size increases the probability because $\sigma_{\bar{x}}$ decreases as $n$ increases. C. Increasing the sample size increases the probability because $\sigma_{\bar{x}}$ increases as $n$ increases. D. Increasing the sample size decreases the probability because $\sigma_{\bar{x}}$ increases as $n$ increases.

Step 1 ：Given that the mean reading speed is 92 words per minute (wpm) and the standard deviation is 10 wpm, we are asked to find the number of samples we would expect to have a sample mean reading rate of less than 98 words per minute, given that 100 different samples of 20 students each were chosen.

Step 2 ：To solve this, we first calculate the z-score for 98 words per minute. The z-score is calculated as \( (X - \mu) / \sigma \), where X is the value we're interested in (98 wpm), \(\mu\) is the mean (92 wpm), and \(\sigma\) is the standard deviation (10 wpm).

Step 3 ：However, since we're dealing with a sample mean, we need to adjust the standard deviation by dividing it by the square root of the sample size (20). This adjusted standard deviation is known as the standard error.

Step 4 ：After calculating the z-score, we can use the standard normal distribution table to find the corresponding probability. This probability represents the proportion of samples we would expect to have a sample mean reading rate of less than 98 words per minute.

Step 5 ：Multiplying this probability by the total number of samples (100) will give us the expected number of such samples.

Step 6 ：We find that we would expect approximately 99.64 samples to have a sample mean reading rate of less than 98 words per minute. However, since we can't have a fraction of a sample, we should round this number to the nearest whole number.

Step 7 ：Final Answer: \(\boxed{100}\). If 100 different samples of \(n=20\) students were chosen from this population, we would expect 100 samples to have a sample mean reading rate of less than 98 words per minute.