Step 1 :Given that the population mean reading speed is 92 words per minute (wpm), the population standard deviation is 10 wpm, the sample size is 19 students, and the sample mean reading speed is 94.9 wpm.
Step 2 :Calculate the standard error, which is the population standard deviation divided by the square root of the sample size. The standard error is approximately \(2.294 \, \text{wpm}\).
Step 3 :Calculate the z-score, which is the sample mean minus the population mean, divided by the standard error. The z-score is approximately 1.264.
Step 4 :Use a z-table to find the probability associated with the z-score. The probability of obtaining a result of 94.9 wpm or more is approximately 0.1031 or 10.31%.
Step 5 :This means that we would expect a mean reading rate of 94.9 or higher from a population whose mean reading rate is 92 in about 10.31 of every 100 random samples of size 19 students.
Step 6 :Since the probability is not very small, it suggests that the new reading program is not abundantly more effective than the old program.
Step 7 :The final answer is \(\boxed{0.1031}\)