The reading speed of second grade students in a large city is approximately normal, with a mean of 89 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts $(a)$ through $(f)$. (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 21 second grade students was $91.4 \mathrm{wpm}$. 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 $91.4 \mathrm{wpm}$ is not unusual since the probability of obtaining a result of $91.4 \mathrm{wpm}$ or more is 0.1357 . This means that we would expect a mean reading rate of 91.4 or higher from a population whose mean reading rate is 89 in 14 of every 100 random samples of size $n=21$ students. The new program is not abundantly more effective than the old program. $\mathrm{X}_{\mathrm{B}}$, A mean reading rate of $91.4 \mathrm{wpm}$ is unusual since the probability of obtaining a result of 91.4 wpm or more is . This means that we would expect a mean reading rate of 91.4 or higher from a population whose mean reading rate is 89 in of every 100 random samples of size $n=21$ students. The new program is abundantly more effective than the old program. (f) There is a $5 \%$ chance that the mean reading speed of a random sample of 22 second grade students will exceed what value? There is a $5 \%$ chance that the mean reading speed of a random sample of 22 second grade students will exceed $\square$ wpm. (Round to two decimal places as needed.)
Solution
Step 1 ：The problem is asking for the value of reading speed that only 5% of a sample of 22 students would exceed. This is a question about the upper tail of the normal distribution.
Step 2 ：We can use the z-score formula to find this value. The z-score formula is: \(Z = \frac{(X - \mu)}{(\sigma / \sqrt{n})}\) where X is the value we're looking for, \(\mu\) is the mean, \(\sigma\) is the standard deviation, and n is the sample size.
Step 3 ：We know that the z-score corresponding to the upper 5% of the normal distribution is approximately 1.645. We can rearrange the formula to solve for X: \(X = Z * (\sigma / \sqrt{n}) + \mu\)
Step 4 ：We can plug in the values we know: Z = 1.645, \(\mu = 89\), \(\sigma = 10\), and n = 22.
Step 5 ：Substituting the values into the formula, we get \(X = 1.645 * (10 / \sqrt{22}) + 89\)
Step 6 ：Solving the equation, we get \(X = 92.5071517840498\)
Step 7 ：Rounding to two decimal places, we get \(X = 92.51\)
Step 8 ：Final Answer: There is a 5% chance that the mean reading speed of a random sample of 22 second grade students will exceed \(\boxed{92.51}\) wpm.
