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).
B. Increasing the sample size increases the probability becáuse $\sigma_{\bar{x}}$ 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.
(e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 19 second grade students was $94.9 \mathrm{wpm}$. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within your choice.
(Type integers or decimals rounded to four decimal places as needed.)
A. A mean reading rate of $94.9 \mathrm{wpm}$ is unusual since the probability of obtaining a result of $94.9 \mathrm{wpm}$ or more is $\square$. 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 $\square$ of every 100 random samples of size $n=19$ students. The new program is abundantly more effective than the old program.
B. A mean reading rate of $94.9 \mathrm{wpm}$ is nar unusual since the probability of obtaining a result of $94.9 \mathrm{wpm}$ or more is $\square$. 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 $\square$ of every 100 random samples of size $n=19$ students. The new program is not abundantly more effective than the old program.
Answer
The final answer is \(\boxed{0.1031}\)
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 \(\boxed{0.1031}\)
