Questions: The reading speed of second grade students in a large city is approximately normal, with a mean of 91 words per minute (wpm) and a standard deviation of 10 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), (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 93.4 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 93.4 wpm is not unusual since the probability of obtaining a result of 93.4 wpm or more is This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in of every 100 random samples of size n=19 students. The new program is not abundantly more effective than the old program. B. A mean reading rate of 93.4 wpm is unusual since the probability of obtaining a result of 93.4 wpm or more is 1. This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in of every 100 random samples of size n=19 students. The new program is abundantly more effective than the old program.

The reading speed of second grade students in a large city is approximately normal, with a mean of 91 words per minute (wpm) and a standard deviation of 10 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), (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 93.4 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 93.4 wpm is not unusual since the probability of obtaining a result of 93.4 wpm or more is This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in of every 100 random samples of size n=19 students. The new program is not abundantly more effective than the old program. B. A mean reading rate of 93.4 wpm is unusual since the probability of obtaining a result of 93.4 wpm or more is 1. This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in of every 100 random samples of size n=19 students. The new program is abundantly more effective than the old program.
Transcript text: The reading speed of second grade students in a large city is approximately normal, with a mean of 91 words per minute (wpm) and a standard deviation of 10 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), (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 93.4 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 93.4 wpm is not unusual since the probability of obtaining a result of 93.4 wpm or more is $\square$ This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in $\square$ of every 100 random samples of size $n=19$ students. The new program is not abundantly more effective than the old program. B. A mean reading rate of 93.4 wpm is unusual since the probability of obtaining a result of 93.4 wpm or more is $\square$ 1. This means that we would expect a mean reading rate of 93.4 or higher from a population whose mean reading rate is 91 in $\square$ of every 100 random samples of size $n=19$ students. The new program is abundantly more effective than the old program.
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Solution

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Solution Steps

Step 1: Calculate the Z-Score

To determine how unusual the sample mean of \( 93.4 \) wpm is, we first calculate the Z-score using the formula:

\[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]

where:

  • \( \bar{X} = 93.4 \) (sample mean)
  • \( \mu = 91 \) (population mean)
  • \( \sigma = 10 \) (population standard deviation)
  • \( n = 19 \) (sample size)

Calculating the Z-score gives us:

\[ Z = \frac{93.4 - 91}{10 / \sqrt{19}} \approx 1.0461 \]

Step 2: Calculate the Probability

Next, we find the probability of obtaining a sample mean of \( 93.4 \) wpm or more. This is done using the standard normal distribution:

\[ P(Z \geq 1.0461) = 1 - P(Z < 1.0461) = 1 - \Phi(1.0461) \approx 0.1477 \]

Step 3: Interpret the Probability

The calculated probability \( P \) indicates that:

\[ P = 0.1477 \]

This means that we would expect to obtain a mean reading rate of \( 93.4 \) wpm or higher from a population whose mean reading rate is \( 91 \) in approximately \( 15 \) out of every \( 100 \) random samples of size \( n = 19 \) students.

Step 4: Draw a Conclusion

Since the probability \( P = 0.1477 \) is greater than \( 0.05 \), we conclude that a mean reading rate of \( 93.4 \) wpm is not unusual. Therefore, we can state:

A mean reading rate of \( 93.4 \) wpm is not unusual since the probability of obtaining a result of \( 93.4 \) wpm or more is \( 0.1477 \). The new program is not abundantly more effective than the old program.

Final Answer

The answer is A.

\(\boxed{A}\)

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