There are numerous ways to represent this data visually and numerically, the first graph I did was the stem and leaf plot where you can see the frequency much clearer since the original data numbers are shown as in compare like a histogram where the frequency is easy to see, the original data points have been lost. For an example the grades, there are 13 values that were in the seventies, but is hard to determine from the histogram what those values might have been. In the stem and leaf plot shows the original values way easier, which is perfect to show to younger students because the horizontal leaves in the stem and leaf plot is the same to the vertical bars in the histogram, and the leaves have the heights that equal the numbers in the …show more content…
In the box and whisker, the median is 78, almost 80, and that fifty percent of the students scored between 67 and 86 points, and seeing the plot is skewed to the left means that there are more values in the highs rather the lower numbers. The histogram and stem and leaf plot also shows how students passed the test by seeing the median and mode on the graph.
The assigning of the scores I decided to have five letter grades which were, A, B, C, D, and F. Letter grade A is 90-100, B is 80-89, C is 70-79, D is 60-69, and F is 59-0. The reason I chose this grade assignment is because I wanted to know how many students have “passed” the exam beyond average or below. The average is a 70-79 and having five categories can show which students actually know the subject or need to work on it, whether over the average or under the average.
Describing the “middle” of the data is that the central tendency by using the mean, median, or mode. The mean gives us the average, the median gives us which is the number in the middle of the set, while mode show us the most frequently occurring value in a data set. The box and whisker and histogram, it’s better to use the median to describe the middle because since the mean is less than the median, that’s why it’s skewed because the high scores influence the direction of the graph. In this case, the median may be a better option to represent the typical grade on the test.
To describe the overall
This is a histogram were the tail goes to the right, it means the average is larger than the median.
Letter Grade: A = 90 - 100%; B = 80 - 89%; C = 70 - 79%; D = 60 - 69%; F = 0 – 59%.
1. By hand, compute the mean, median, and mode for the following set of 40 reading scores:
The test is administered in the spring and the results provide students, teachers, and parents with an objective view of the student’s strengths and weaknesses. The listening and speaking scores are combined to obtain an overall scale score, as are the reading and writing scores. The overall performance is determined by the scores in each combined area. A conversion chart is provided in each grade band, to convert the raw score to a scale score. Then depending on where the scale scores land will give you the placement level that the student is at. To move from one level to another both component scores must land in the upper level range. A student, who advances in only one
The system should be able to produce a final score out of ten for each student.
The data that I have is accurate because by looking at the histogram the graphs are perfectly skewed and, the data make sense, if I compare the data to the median and the IQR. And to find IQR subtract Quartile 3 from Quartile 1. The purpose of using mean value to see if the graph is symmetric or not but, in my case I do not have symmetrical graph or Boxplot, because it is sensitive to outliers. I choose to use the IQR instead of the standard deviation because it shows what is below the 1st quartile and above the 3rd quartile. To summarize, median and IQR are the best option because they resistant to outliers.
Mean would be the most appropriate measure of central tendency to describe this data. This is because the mean is the average of all scores in the data set. If Dr. Williams were to graph the data into a bell shaped distribution, then the mean would be in the center where most of the scores are located. The mean is calculated using all information of the data set, and is the best score to use if you want to predict an individual score.
Based on the given sample of student test scores of 50, 60, 74, 83, 83, 90, 90, 92, and 95 after rearranging them from least to greatest. As the mean is based on the average of sum, the average of this sample is 79.67 or 80. The mode refers to numbers that appear the most in a sequence and in this case 83 and 90 both appear twice. Range calculates the difference between the largest and smallest number, which are 95 and 50 which have a difference of 45. The variance is the difference between the sum of squares divided by the sample size, which is the number in the sample minus one (Hansen & Myers, 2012), meaning it takes each number of the set and subtracts
An additional method to determine an outlier is to consider the distribution of the outlier to the median as opposed to the mean. Based on this principle, a value is considered an outlier if “positioned 1.5 times the value of the 75th percentile in a distribution” (Sylvia & Terhaar, 2014). In the case study, the boxplot figure displays values above and below the 75th percentile or the whiskers (Sylvia & Terhaar, 2014). However, the principal investigator of the study determined these values should be included as they provide meaning to the final results. In some situations when the outliers cause a skew in the distribution, other solutions or techniques should be pursued for analysis(Sylvia & Terhaar,
Now that the data are sorted, one result is the range of the data from 0 to 20, i.e. 21. The median is the middle observation in an odd number of sorted data and halfway between the two center-most points
In picture B show the result central tendency of age. Min=32.00, Q1=59.75, Median=71.00, Mean=68.25, Q3=80.25 and Max=92.00. When I interpret this result. I compare two values (median-Q1 and Q3-median). The value from Q3-median is more than value from median-Q1 that mean this graph is negative skew, correlate with picture C.
Students will receive a final exam score of 1-5, derived from their performance on both the through-course assessment and the end-of-course
To obtain direct measures of ROM, I would perform the goniometer test and based on her findings, I believe that indirect methods would be useful to assess the flexibility of her lower back and hamstrings. The Sit- and - Reach Test is useful to assess the “static flexibility of lower back and hamstring muscles” (Heyward, p. 273, 2010). The Standard Sit-and -Reach test evaluates lower back and hamstring flexibility. I would perform the Back Saver Sit-and –Reach test because it relieves some of the discomfort when measuring flexibility of the hamstring muscles. Also, the skin distraction test is useful in assessing low back flexibility. Furthermore, based on the results of the flexibility tests, I will design a flexibility program for her and
Interpretation: The boxplot displays the behavior score distributions by grade. The behavior scores differ significantly by grade level. The lower high school has a higher average median score than both the middle school and upper high school. The lower high school has the smallest variance in scores while the upper high school has the largest variance in scores. There were a disproportionate number of middle school observations when compared to the high school data.
Majority of the students measured were 70 inches or less. ¾ of those surveyed measured 70 inches or less, the other ¼ of those surveyed measured above 70inches. The shape of this graph would be left skewed, most of the data is on the right side of the distribution.