Mark is interested in seeing how outgoing his residents are so he distributes a survey on extraversion to them. Mark is only interested in his own residents and is able to give surveys to all of them, so the data he collects represents a population.
x (extraversion score)
10
10
9
8
8
7
5
5
5
4
1) For the population Mark has data from, please calculate the following:
a) Mean:
b) Sum of Squares:
c) Variance:
d) Standard Deviation:
2) As you can see from the data, one of Mark’s residents received an extraversion score of 9. Please convert that resident’s score to a z-score (using the mean and standard deviation you found previously). What is the z-score for this resident?
3) Now we want to convert this resident’s score to a new distribution where the mean is 100 and the standard deviation is 10. What is this resident’s new raw score?
a)
mean =7.10\
b)sum of square =44.90
c)variance =4.99
d)Std deviaiton =2.23
2)
as z score =(X-mean)/std deviaiton
hence
S.No | X | z score | |
1 | 10.00 | 1.30 | |
2 | 10.00 | 1.30 | |
3 | 9.00 | 0.85 | |
4 | 8.00 | 0.40 | |
5 | 8.00 | 0.40 | |
6 | 7.00 | -0.04 | |
7 | 5.00 | -0.94 | |
8 | 5.00 | -0.94 | |
9 | 5.00 | -0.94 | |
10 | 4.00 | -1.39 |
3)
for new raw score =mean +z*std deviaion =100+z*10
S.No | X | z score | new raw score | |
1 | 10.00 | 1.30 | 113 | |
2 | 10.00 | 1.30 | 113 | |
3 | 9.00 | 0.85 | 108.5 | |
4 | 8.00 | 0.40 | 104 | |
5 | 8.00 | 0.40 | 104 | |
6 | 7.00 | -0.04 | 99.6 | |
7 | 5.00 | -0.94 | 90.6 | |
8 | 5.00 | -0.94 | 90.6 | |
9 | 5.00 | -0.94 | 90.6 | |
10 | 4.00 | -1.39 | 86.1 |
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