Problem 1
The performances of a group of interns are evaluated by their supervisors at the end of their
internships.
Their scores are: 55, 47, 62, 27, 50, 49, 66, 53, 50, 44, 63, 59.
Complete the calculations below using this data. Show all of your work and clearly label each of
your calculations.
the meanthe medianthe rangethe standard deviationthe variance
Problem 2The 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. What is the z score equivalent of ANGST = 81? What is the probability that
someone selected at random will score 81 or lower? What percentage of all trainees will score
between 60 and 75?
Solution
2.
sample size,n=10
sample mean,u=59.5
sample standard deviation,s=11.098
x=81
z-score = (x-u)/(s/sqrt(n)) = 6.126
P(X<=81) = P(Z<=6.126)
= 1
x1=60 z1=(x1-u)/(s/sqrt(n)) = 0.14
x2=75 z2=(x2-u)/(s/sqrt(n)) = 4.42
P(60.
Problem 1 The performances of a group of interns are evaluat.pdf
1. Problem 1
The performances of a group of interns are evaluated by their supervisors at the end of their
internships.
Their scores are: 55, 47, 62, 27, 50, 49, 66, 53, 50, 44, 63, 59.
Complete the calculations below using this data. Show all of your work and clearly label each of
your calculations.
the meanthe medianthe rangethe standard deviationthe variance
Problem 2The 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. What is the z score equivalent of ANGST = 81? What is the probability that
someone selected at random will score 81 or lower? What percentage of all trainees will score
between 60 and 75?
Solution
2.
sample size,n=10
sample mean,u=59.5
sample standard deviation,s=11.098
x=81
z-score = (x-u)/(s/sqrt(n)) = 6.126
