Question

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). ​

(a) What is the probability a randomly selected student in the city will read more than 96 words per​ minute? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice.

A. If 100 different students were chosen from this​ population, we would expect nothing to read more than 96 words per minute.

B. If 100 different students were chosen from this​ population, we would expect nothing to read less than 96 words per minute.

C. If 100 different students were chosen from this​ population, we would expect nothing to read exactly 96 words per minute. ​

(b) What is the probability that a random sample of 12 second grade students from the city results in a mean reading rate of more than 96 words per​ minute? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice.

A. If 100 independent samples of nequals12 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of less than 96 words per minute.

B. If 100 independent samples of nequals12 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of exactly 96 words per minute.

C. If 100 independent samples of nequals12 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of more than 96 words per minute. ​

(c) What is the probability that a random sample of 24 second grade students from the city results in a mean reading rate of more than 96 words per​ minute? The probability is nothing. ​(Round to four decimal places as​ needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice.

A. If 100 independent samples of nequals24 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of more than 96 words per minute.

B. If 100 independent samples of nequals24 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of less than 96 words per minute.

C. If 100 independent samples of nequals24 students were chosen from this​ population, we would expect nothing ​sample(s) to have a sample mean reading rate of exactly 96 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 Subscript x overbar decreases as n increases.

B. Increasing the sample size increases the probability because sigma Subscript x overbar decreases as n increases.

C. Increasing the sample size increases the probability because sigma Subscript x overbar increases as n increases.

D. Increasing the sample size decreases the probability because sigma Subscript x overbar 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 21 second grade students was 93.7 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 93.7 wpm is unusual since the probability of obtaining a result of 93.7 wpm or more is nothing. This means that we would expect a mean reading rate of 93.7 or higher from a population whose mean reading rate is 91 in nothing of every 100 random samples of size nequals21 students. The new program is abundantly more effective than the old program.

B. A mean reading rate of 93.7 wpm is not unusual since the probability of obtaining a result of 93.7 wpm or more is nothing. This means that we would expect a mean reading rate of 93.7 or higher from a population whose mean reading rate is 91 in nothing of every 100 random samples of size nequals21 students. The new program is not abundantly more effective than the old program. ​

(f) There is a​ 5% chance that the mean reading speed of a random sample of 21 second grade students will exceed what​ value? There is a​ 5% chance that the mean reading speed of a random sample of 21 second grade students will exceed nothing wpm. ​(Round to two decimal places as​ needed.)

Answer 1

Answer 2

(a) The probability that a randomly selected student in the city will read more than 96 words per minute can be calculated using the standard normal distribution.

First, we calculate the z-score using the formula: z = (x - μ) / σ

where x is the value (96 words per minute), μ is the mean (91 words per minute), and σ is the standard deviation (10 words per minute).

z = (96 - 91) / 10 z = 0.5

Next, we find the probability using a standard normal distribution table or calculator. The probability associated with a z-score of 0.5 is approximately 0.6915.

Therefore, the probability that a randomly selected student in the city will read more than 96 words per minute is 0.6915.

Interpretation: If 100 different students were chosen from this population, we would expect approximately 69.15 of them to read more than 96 words per minute.

(b) The probability that a random sample of 12 second-grade students from the city results in a mean reading rate of more than 96 words per minute can be calculated using the sampling distribution of the sample mean.

Since the sample size (n) is relatively small (12) and the population standard deviation (σ) is known, we can use the t-distribution. However, since we want to find the probability of a mean reading rate greater than 96, which is not at the center of the distribution, we can approximate it using the normal distribution.

Using the central limit theorem, the sampling distribution of the sample mean follows an approximately normal distribution with a mean equal to the population mean (μ) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σ/sqrt(n)).

The z-score can be calculated as: z = (x - μ) / (σ / sqrt(n)) z = (96 - 91) / (10 / sqrt(12)) z = 2.68

The probability associated with a z-score of 2.68 can be found using a standard normal distribution table or calculator. However, since the probability is stated as "nothing" (which is likely an error), we cannot provide a specific probability value in this case.

Interpretation: If 100 independent samples of n=12 students were chosen from this population, we cannot determine the exact number of samples that would have a sample mean reading rate of more than 96 words per minute based on the given information.

(c) Similarly, the probability that a random sample of 24 second-grade students from the city results in a mean reading rate of more than 96 words per minute can be calculated using the sampling distribution of the sample mean.

Using the same approach as in part (b), we calculate the z-score: z = (96 - 91) / (10 / sqrt(24)) z = 2.12

Again, the probability associated with a z-score of 2.12 can be found using a standard normal distribution table or calculator. However, since the probability is stated as "nothing," we cannot provide a specific probability value in this case.

Interpretation: If 100 independent samples of n=24 students were chosen from this population, we cannot determine the exact number of samples that would have a sample mean reading rate of more than 96 words per minute based on the given information.

(d) Increasing the sample size has an effect on the probability of obtaining a sample mean. As the sample size (n) increases, the standard deviation of the sample mean (σ_x̄) decreases. This means that the sample mean becomes a more precise estimate of the population mean, resulting in a narrower distribution of sample means.