Asked by joane
1) Entry to a certain University is determined by a national test. The scores on this
test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score
better than at least 70% of the students who took the test. Tom takes the test and
scores 585. Will he be admitted to this university?
^ I already did this question and I got 80.23%
but I need help with the other questions that go with this :
2) For the same test, consider Sarah, whose score is 683. What is her
a) Z score
b) T score
c) Percentile rank
d) What percentage of people scored between Sarah and the mean? How many
people were ahead of her? (Clue: use the Z table).
e) If the mean on the test were 550 and the Sd = 50, what would be Tom and
Sarah’s scores? Would either of them qualify for the University now?
test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score
better than at least 70% of the students who took the test. Tom takes the test and
scores 585. Will he be admitted to this university?
^ I already did this question and I got 80.23%
but I need help with the other questions that go with this :
2) For the same test, consider Sarah, whose score is 683. What is her
a) Z score
b) T score
c) Percentile rank
d) What percentage of people scored between Sarah and the mean? How many
people were ahead of her? (Clue: use the Z table).
e) If the mean on the test were 550 and the Sd = 50, what would be Tom and
Sarah’s scores? Would either of them qualify for the University now?
Answers
Answered by
PsyDAG
I did not check #1.
2. Z = (score-mean)/SD
Z = (683-500)/100 (calculate)
T = 50 + 10Z
Percentile rank = proportion ≤ score
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores for c, d and e.
2. Z = (score-mean)/SD
Z = (683-500)/100 (calculate)
T = 50 + 10Z
Percentile rank = proportion ≤ score
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores for c, d and e.
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