The reading speed of second grade students in a large city is approximately norm

Question

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). B. Increasing the sample size increases the probability because sigma^-_x increases as n increases. C. Increasing the sample size decreases the probability because sigma^-_x decreases as n increases. D. Increasing the sample size increases the probability because sigma^-_x decreases 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 18 second grade students was 91.5 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 91.5 wpm is unusual since the probability of obtaining a result of 91.5 wpm or more is This means that we would expect a mean reading rate of 91.5 or higher from a population whose mean reading rate is 89 in of every 100 random samples of size n = 18 students. The new program is abundantly more effective than the old program. B. A mean reading rate of 91.5 wpm is not unusual since the probability of obtaining a result of 91.5 wpm or more is This means that we would expect a mean reading rate of 91.5 or higher from a population whose mean reading rate is 89 in of every 100 random samples of size n = 18 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 20 second grade students will exceed what value? There is a 5% chance that the mean reading speed of a random sample of 20 second grade students will exceed wpm. (Round to two decimal places as needed.)

Explanation / Answer

(e) Mean = 89 Words per minute = 89 WPM

Standard deviation = 10 Words per minute = 10 WPM

Sample size = 18

Sample mean = 91.5 wpm

Pr ( X > 91.5 wpm; 89; 10/ sqrt (18) ) = 1 - Pr( X < 91.5 wpm; 89; 10/ sqrt (18) )

Z = (91.5 - 89)/ (10/ sqrt (18) ) = 2.5/ 2.357 = 1.06

Pr(Z>1.06) = 1 - 0.8554 = 0.1446

A mean reading rate of 91.5 wpm is not unusual since the probability of obtaing a result of 91.5 wpm or more is 0.1446. This means that we would expect a mean reading rate of 91.5 or higher from a population whose mean reading rate is 89 in 14 of every 100 random samples of size n = 18 students. The new program is not abundantly more effecive than the old program.

(f) sample size = 20

Pr (x > X wpm; 89; 10/ sqrt (20)) = 0.05

so Z = 1.645

( X - 89)/ (10/ sqrt (20)) = 1.645

X = 89 + 1.645 * (10/ sqrt (20))

X = 89 + 1.645 * 2.236

X = 92.68 WPM

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