Question 772066: The Anxiety General Stress Test (ANGST) has been designed to gauge the level of psychological stress management trainees experience when they are under pressure.
For a random sample of trainees the scores are as follows: 47, 49, 53, 53, 54, 58, 61, 64, 75, 81.
Complete the calculations below using this data. Show all of your work and clearly label each of your calculations.
a.What is the z score equivalent of ANGST = 81?
b.What is the probability that someone selected at random will score 81 or lower?
c.What percentage of all trainees will score between 60 and 75?
OK for answer (a) I have:
z = (81 - 59.5)/11.098
z = 21.5/11.098
z = 1.937 or 1.94 (rounded up)
is that correct?
Answer by John10(297)   (Show Source):
You can put this solution on YOUR website! Hi,
For this problem, you will need to find the MEAN and STANDARD DEVIATION from the data.
After you found both, you will able to solve part (a) to part (c):
a) z = (81 - mean)/ s. d = ?
b) P(X <= 81) = P( z < result of part a) = ?
Note: X is variable
c) P( 60 < x < 75)
You will find
z(60) and z(75) the same process in part (a) and part (b)
z(60) = (60 - mean)/ s.d ?
P(z(60) < z < z(75) = ?
z(75) = (75 - mean)/s.d = ?
Hope it helps you!
John10 (john100185@yahoo.com)
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