4. What is the range (sxs _) of the data values that would allow 55% of the data to fall within the mean observing symmetry? Draw the curve and show your complete solution. The Normal Distribution Activity For Data Set 2 Ar4.10; 3. Mean = 2.00 Variance = 4.5263 Standard Deviation = 2.1275 Ag 0. G E NEF or e t ns 7. Sketch the second bell curve by showing the mean and standard deviation. EN .26 TO PAL NE N teNE hart 1.6 15: -t.0 ** 10. 1945 - L6S15 x-t.10 The range of data values that will allow 55% of the data to fall within the mean of symmetry is 7.0625 and 10.3375 5. What is the variance of this data set? o? = (2.1546)? g? = 4.6423 .af e Given that the standard deviation is 2.1546 and it is the square root of the variance, 8. What is the data value (raw score) associated with a standard score of 2.08? Determine the percentage greater than this score. Draw the curve and show your complete solution. the value of the variance is 4.6423. 6. Are there any outliers with this data set? (Outliers refer to data points that lie beyond 3 standard deviation from the mean). A.cecino vers 2.5 There are no outliers with this data set since all values in the data set do not go beyond the three standard deviations from the mean. t. ceree t e er O 2.00 The percentage that is greater than the raw score of 6.4252 is 98.12%. 9. What is the data value (raw score) associated with a standard score of -1.3? Determine the percentage less than this score. Draw the curve and show your complete solution. -1- 16t x- * 0.9032 10. 81%. LE N ees The percentage that is less than the raw score of -0.7658 is 90.32%.
4. What is the range (sxs _) of the data values that would allow 55% of the data to fall within the mean observing symmetry? Draw the curve and show your complete solution. The Normal Distribution Activity For Data Set 2 Ar4.10; 3. Mean = 2.00 Variance = 4.5263 Standard Deviation = 2.1275 Ag 0. G E NEF or e t ns 7. Sketch the second bell curve by showing the mean and standard deviation. EN .26 TO PAL NE N teNE hart 1.6 15: -t.0 ** 10. 1945 - L6S15 x-t.10 The range of data values that will allow 55% of the data to fall within the mean of symmetry is 7.0625 and 10.3375 5. What is the variance of this data set? o? = (2.1546)? g? = 4.6423 .af e Given that the standard deviation is 2.1546 and it is the square root of the variance, 8. What is the data value (raw score) associated with a standard score of 2.08? Determine the percentage greater than this score. Draw the curve and show your complete solution. the value of the variance is 4.6423. 6. Are there any outliers with this data set? (Outliers refer to data points that lie beyond 3 standard deviation from the mean). A.cecino vers 2.5 There are no outliers with this data set since all values in the data set do not go beyond the three standard deviations from the mean. t. ceree t e er O 2.00 The percentage that is greater than the raw score of 6.4252 is 98.12%. 9. What is the data value (raw score) associated with a standard score of -1.3? Determine the percentage less than this score. Draw the curve and show your complete solution. -1- 16t x- * 0.9032 10. 81%. LE N ees The percentage that is less than the raw score of -0.7658 is 90.32%.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Answer 7 8 9 handwritten
Transcribed Image Text:4. What is the range (_<x <_) of the data values that would allow 55% of the data to fall
The Normal Distribution Activity
within the mean observing symmetry? Draw the curve and show your complete solution.
For Data Set 2
A : 55 %.
A, :-0. 2750 jz- 0.76
Мean 3D
2.00
Variance =
4.5263
Standard Deviation =
2.1275
: 0. 5500
Ag : 0. 2750 ; 2:0.76
THE RANCE OF THE DATA UALUES
THA1 WILL ALLOW 55%. OF THE DATA
TO FALL WITHIN THE MEAN OBSERUING
SYM METRY I5 7.OG26 S X 4 10. 3375
7. Sketch the second bell curve by showing the mean and standard deviation.
WHEN 2 0. 96
WHEN 2: -0.7G
0.76 * - 8.70
-0. 7G = 7-P.70
2-154G
2.154G
-0. 1G
0. 1G
- 1.63 15 =x -P.70
1.6 315 : * - t.70
*e 10. 39 45
** 1.0625
The range of data values that will allow 55% of the data to fall within the mean of symmetry is
7.0625 and 10.3375
5. What is the variance of this data set?
o2 =
(2.1546)?
o2 = 4.6423
Given that the standard deviation is 2.1546 and it is the square root of the variance,
8. What is the data value (raw score) associated with a standard score of 2.08? Determine the
percentage greater than this score. Draw the curve and show your complete solution.
the value of the variance is 4.6423.
6. Are there any outliers with this data set? (Outliers refer to data points that lie beyond 3
standard deviation from the mean).
2.08 - * - 2
Aocze 2.08 = 0. 4e12
2.1275
Az >1.0r : 0.5t 0. NRI2
There are no outliers with this data set since all values in the data set do not go beyond
4.4252 : X - 2
the three standard deviations from the mean.
X - 6. 4252
ge 12 %% CREATER THAN CAW SCORE
2.08
The percentage that is greater than the raw score of 6.4252 is 98.12%.
9. What is the data value (raw score) associated with a standard score of -1.3? Determine the
percentage less than this score. Draw the curve and show your complete solution.
-1-3: x - 2
A-1-3 <Zc0= o- YO82
2.1245
AZL-1.3 = 0.5 + 0. 4032
- 2. 7658 = x-2
0.9032
*: -0. 1658
90. 81 °%. LESS THAN RAW SCORE
- 1.3
The percentage that is less than the raw score of -0.7658 is 90.32%.
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