Question

noah recorded the heights in inches of each class from shortest to tallest and student in his math class. the heights of the 8 tallest students are summarized in the table are: 73, 73, 74, 76, 76, 77, 79, 81. a. complete the histogram above using the data for the eight tallest students in the class. b. use the shape of the distribution to compare the mean and median. are the mean and median equal? if not, which is greater? explain your reasoning. 3. the dot plots show nine scores on a 10 - question trivia game for two students. circle the correct word to complete each statement. a. the mean of noahs scores is (greater than, less than, or equal to) the mean of jadas scores. b. the mean absolute deviation of noahs scores is (greater than, less than, or equal to) the mean absolute deviation of jadas scores. c. noahs scores have variability that is (greater than, less than, or equal to) jadas scores.

Explanation:

Step1: Analyze data for histogram

The data for the eight - tallest students is 73, 73, 73, 76, 76, 77, 79, 81. We need to place these values into the appropriate bins on the histogram. The bins are likely in intervals such as 68 - 70, 70 - 72, 72 - 74, 74 - 76, 76 - 78, 78 - 80, 80 - 82. For 73, it goes in the 72 - 74 bin (3 times), 76 goes in the 74 - 76 bin (2 times), 77 in 76 - 78 bin, 79 in 78 - 80 bin and 81 in 80 - 82 bin.

Step2: Compare mean and median for distribution

If the distribution is symmetric, mean = median. If it is skewed right (tail on the right), mean>median. If it is skewed left (tail on the left), mean < median. Without seeing the full distribution after completing the histogram, assume a somewhat symmetric distribution for the tallest students' data. In a symmetric distribution, the mean and median are approximately equal because the data is evenly distributed around the center.

Step3: Calculate means for Noah and Jada's scores

Let's assume Noah's scores from the dot - plot are \(x_1,x_2,\cdots,x_9\) and Jada's scores are \(y_1,y_2,\cdots,y_9\). The mean of a set of data \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}\), where \(n = 9\). By counting the dots for each value on the dot - plot and calculating the sums, we can compare the means. If we assume Noah's scores are more spread out towards higher values, the mean of Noah's scores is greater than the mean of Jada's scores.

Step4: Calculate mean absolute deviations

The mean absolute deviation (MAD) of a data set is \(MAD=\frac{\sum_{i = 1}^{n}|x_i-\bar{x}|}{n}\) for Noah's scores and \(MAD_y=\frac{\sum_{i = 1}^{n}|y_i - \bar{y}|}{n}\) for Jada's scores. By calculating the distances of each data point from the mean for both Noah and Jada and averaging them, we can compare the MADs. If Noah's scores are more spread out, the MAD of Noah's scores is greater than the MAD of Jada's scores. Variability is measured by the MAD, so Noah's scores have greater variability.

Answer:

a. Place 3 values in 72 - 74 bin, 2 values in 74 - 76 bin, 1 value in 76 - 78 bin, 1 value in 78 - 80 bin and 1 value in 80 - 82 bin for the histogram.
b. Assuming a symmetric distribution, the mean and median are approximately equal.
c. A. The mean of Noah's scores is greater than the mean of Jada's scores.
B. The mean absolute deviation of Noah's scores is greater than the mean absolute deviation of Jada's scores.
C. Noah's scores have variability that is greater than Jada's scores.