Standard normal variate is
Z = (X - Mean) / SD
0.5 = (X - 84) / 10
X = 0.5*10 + 84
= 89
Your grade on the test is 89
Quiz: Midterm & help file The mean grade on a French test was 84 with a...
Suppose the mean grade for a statistics midterm exam was 75, with a standard deviation of 10. Assume that your grades were normally distributed. [ 40 pts. Total] a. What percentage of students received at least an 90? [7.5 pts] b. What percentage of students received an 84? [5 pts.] c. What percentage of students received a grade between 60 and 79? [10 pts.] d. What percentage of students received a grade less than 70? [7.5 pts.] e. If 2...
THINKING/COMMUNICATION 7. A post-secondary class has 600 students. The mean grade on their midterm exam is 70%, with a standard deviation of 15. a. Find the Z-score of a student who scored 80% on his exam. [2 marks] b. Find the percentage of students who failed the exam. [3 marks] c. How many students failed the exam (i.e. scored less than 49%)? [1 mark] d. Your friend claims he scored in the 99th percentile on his exam. What mark must...
In a large class, there were 205 students who wrote both the midterm and the final exam. The standard deviation of the midterm grades was 15, and that of the final exam was 18. The correlation between the grades on the midterm and the final was 0.58 Based on the least squares regression line fitted to the data of the 205 students, if a student scored 21 points below the mean on the midterm, then how many points below the...
A teacher sets a test for a class of students. The teacher decides to analyse the test results using statistical analysis techniques. Given the table of results below, complete the following tasks: CLASS RESULTS STUDENTS TEST SCORE% Thomas 74 Charles 55 Sarah 81 Mathew 68 James 71 Jessica 74 Daniel 86 Jack 66 Emma 73 Laura 72 Joshua 84 Alice 68 Samantha 70 a). Calculate the Mean of the test scores. b). Calculate the Median of the test scores. c)....
A teacher sets a test for a class of students. The teacher decides to analyze the test results using statistical analysis techniques. Given the table of results below, complete the following tasks: CLASS RESULTS Student Test Score (%) Thomas 74 Charles 55 Sarah 81 Mathew 68 James 71 Jessica 74 Daniel 86 Jack 66 Emma 73 Laura 72 Joshua 84 Alice 68 Samantha 70 a). Calculate the Mean of the test scores. b). Calculate the Median of the test scores....
Suppose that the midterm score of a class is normally distributed with the mean of 68.2 points and the standard deviation of 11.3 points. Answer each question. Sketch the curve of the distribution representing the midterm score. Make sure to mark the mean and three standard deviations to either side of the mean. Find the probability that a randomly selected student has score at most 65.9 To be in the top 20% of the class, you need to score what...
Question Help A student's course grade is based on one midterm that counts as 10% of his final grade, one class project that counts as 20% of his final grade, a set of homework assignments that counts as 50% of his final grade, and a final exam that counts as 20% of his final grade. His midterm score is 77, his project score is 92, his homework score is 77, and his final exam score is 69 What is this...
The marks on a statistics midterm test are normally distributed with a mean of 75 and a standard deviation of 6. What is the probability that a sample of 25 exams has a sample mean between 71 and 73? a. 0.45293 b. 0.0475 c. 0.04793 d. 0.0485 e. 0.04707
After receiving your graded midterm, you are told that the exam had an approximately normal distribution. You are told by your proof that you scored a standard deviation above average. You scored higher than what percent of students?
The average test scores was a 78 with a variance of 16 calculate the Z score for a student who scored an 70. Z = ____ A. 1 standard deviation below the mean B. 0.5 standard deviations above the mean C. 2 standard deviations below the mean D. 2 standard deviations above the mean E. 1 standard deviations above the mean F. 0.5 standard deviations below the mean