Asked by Ana
The mean of Exam 1 is 80%. The variance is 25. How many students of my IS310 class of 200 students have scores between 70% and 90%, assuming that the distribution is bell-shaped?
a.180
b.185
c.190
d.195
I'm not sure how to set up this problem.
a.180
b.185
c.190
d.195
I'm not sure how to set up this problem.
Answers
Answered by
PsyDAG
Z = (score-mean)/SEm
SEm = SD/√n
SD = √variance
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for each and the Z score. Insert into above equation.
SEm = SD/√n
SD = √variance
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for each and the Z score. Insert into above equation.
Answered by
Ana
I'm not allowed to use the chart. I have to calculate it by hand
Answered by
J
So I think the way the professor wanted to answer this is knowing the empirical rule because its normal
65%= 1 SD
95%=2 SD
98%=3 SD
all from the mean
Variance of 25 gives us a Standard deviation of 5.
80% is the mean. So moving 2 SD up gives us 90%
Its normal and the empirical rule states that 2 SD is 95%
So you multiply (95%)x 200= 190
65%= 1 SD
95%=2 SD
98%=3 SD
all from the mean
Variance of 25 gives us a Standard deviation of 5.
80% is the mean. So moving 2 SD up gives us 90%
Its normal and the empirical rule states that 2 SD is 95%
So you multiply (95%)x 200= 190
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