sures the spread or dispersal x2, x3, ..., xn is a collection

Computer Networking: A Top-Down Approach (7th Edition)
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4)
Curve Grades Statisticians use the concepts of mean and standard deviation to
describe a collection of numbers. The mean is the average value of the numbers,
and the standard deviation measures the spread or dispersal of the numbers
about the mean. Formally, if x1, x2, x3, ..., xn is a collection of numbers, then
the mean is
х1 + x2 + х3 +
+ xn
...
m =
n
and the standard deviation is
(x1 – m)2 + (x2 – m)² + (x3 – m)2 +
+ (x4 – m)²
...
S =
n
Transcribed Image Text:4) Curve Grades Statisticians use the concepts of mean and standard deviation to describe a collection of numbers. The mean is the average value of the numbers, and the standard deviation measures the spread or dispersal of the numbers about the mean. Formally, if x1, x2, x3, ..., xn is a collection of numbers, then the mean is х1 + x2 + х3 + + xn ... m = n and the standard deviation is (x1 – m)2 + (x2 – m)² + (x3 – m)2 + + (x4 – m)² ... S = n
The file Scores.txt contains exam scores. The first four lines of the file hold the
numbers 59, 60, 65, and 75. Write a program to calculate the mean and
standard deviation of the exam scores, assign letter grades to each exam score,
ES, as follows, and then display information about the exam scores and the
grades.
Possible outcome:
Number of scores: 14
Average score: 71.0
Standard deviation of scores: 14.42
GRADE DISTRIBUTION AFTER CURVING GRADES.
A: 2
B: 1
C: 6
D: 4
F: 1
ES > m + 1.5s
A
m + .5s < ES < m + 1.5s
m - .5s < ES < m + .5s
C
m = 1.5s < ES < m - .5s
D
ES < m
1.5s
Transcribed Image Text:The file Scores.txt contains exam scores. The first four lines of the file hold the numbers 59, 60, 65, and 75. Write a program to calculate the mean and standard deviation of the exam scores, assign letter grades to each exam score, ES, as follows, and then display information about the exam scores and the grades. Possible outcome: Number of scores: 14 Average score: 71.0 Standard deviation of scores: 14.42 GRADE DISTRIBUTION AFTER CURVING GRADES. A: 2 B: 1 C: 6 D: 4 F: 1 ES > m + 1.5s A m + .5s < ES < m + 1.5s m - .5s < ES < m + .5s C m = 1.5s < ES < m - .5s D ES < m 1.5s
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