The following figure shows the normal distribution with the proportion of the ar

Question

The following figure shows the normal distribution with the proportion of the area under the normal curve contained within one, two, and three standard deviations of the mean. The last proportion on each side, 0.13%, depicts the remaining area under the curve. Specifically, 0.13% of the area under the standard normal distribution is located above z-score values greater than the mean (H) plus three standard deviations (+3a). Also, because the normal distribution is symmetrical, 0.13% of the area under the standard normal distribution is located below z-score values less than the mean (U) minus three standard deviations (-30 34.13% 34.13% 13.59% 13.59% 2.15% 2.15% 0.13% 0.13% 30-20-1o Use the figure to help you answer the following questions The National Assessment of Educational Progress (NAEP) is a nationwide assessment of students' proficiency in nine subjects: mathematics, reading, writing, science, the arts, civics, economics, geography, and U.S. history. The main NAEP assessments are conducted annually on samples of students from grades 4, 8, and 12 In 2002, the reading scores for female students had a mean of 269 with a standard deviation of 33. Assume that these scores are normally distributed with the given mean and standard deviation A score of 203 is below the mean, while a score of 335 is above the mean. This means that the percentage of female students with scores between 203 and 335 is A score of 368 is scores below 368 is above the mean. As a result, the percentage of female students with You can infer that 84.13% of the female students have scores above

Explanation / Answer

A score of 203 is 66 below the mean, while a score of 335 is 66 above the mean. This mean that the percentage of faemales students with scores between 203 and 335 is
P(203<X<335) = P( (203-269) / 33 <Z< (335-269) / 33) = P( -2 < Z < 2) = 0.957 = 95.7%

A score of 368 is 165 aboe the mean. The percentage of female students with scores below 368 is
P(X<368) = P(Z< (368-269)/33) = P(Z < 3) = 0.0013+0.0215+0.1359+0.3413 +0.3413 +0.1359+0.0215 = =0.9987

84.13% of the female students have scores above is 263 - (1) * 33 =230

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