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 PE3.1
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April 28, 2014 at 8:25 pm #13772
Ezra HalleckParticipant1. (40 points) The scores of an exam were normally distributed with a mean of 67 and a standard deviation of 18. The following chart shows percentiles for student grades (e.g., a student who scores better than at least 10% of his or her fellow students but less than 30% will receive a D):
percentile grade
09 F
1029 D
3039 C
4049 C+
5059 B
6069 B
7079 B+
8089 A
9099 A
a. What is the probability that a random score is between 45 and 75? (Hint: do NOT use the percentile chart.)
b. If Sally’s score was at the 35th percentile, what was it numerically and as a letter grade?
c. To get a C, Jack must have what score at a minimum?
d. Jill thought she got an A but instead got a B+. Jill must have scored below what value?May 2, 2014 at 8:29 pm #17946
Aziza R.ParticipantA. P(45>x>75)
4567/18=1.22
7567/18=.44
P(45>x>75)= P(1.22>z>.44)
P(z>.44) – P(z>1.22)
.6700.1112=.5588B. P35%
.6500= z=.39
Formula – m+za
67+.39(18)=74.02 or C
May 2, 2014 at 8:29 pm #17947
Aziza R.ParticipantI wasnt sure how to solve the rest of the problem
May 3, 2014 at 3:07 pm #17948
vivian gomezParticipantA. I go exactly the same as Aziza R.
B. 35%= 0.35 In the Table the closest to 0.35 is 0.3483 which is Z value correspond to – 0.39
To find the X value we use the formula X=67+ 0.39 (18) X= 59.98
Sally got a 59.98 which is a BIf I’m wrong in something please comment
May 3, 2014 at 3:18 pm #17949
vivian gomezParticipantC.
To get a C jack must be on the 30 to 39 Percentile
30 is the lowest
0.30 in the table the closest is 0.2981 which is Z value is 0.53
X=67+0.53(18)
X= 57.46Jack must score a minimum of a 57.46 to get a C
IF you think something is wrong please comment. Lets help each other here
May 3, 2014 at 3:23 pm #17950
vivian gomezParticipantD.
The lowest percentile for an A is 90%0.90= 0.8997 which is a Z value of 1.28
X= 67+1.28(18)= 90.04
Jill socore below the 90 and that why she got a B
May 3, 2014 at 3:54 pm #17951
OruadaParticipant(1a). P(45
4567/18=1.22…Z=.6700
7567/18=0.44…..Z=.1112
.6700.1112=0.558
(1B). P(35)..Z=0.39
67+0.39(18)=74.02…WHICH IS a B+
(1C) P(303067/18=2.05…z=.0202
3967/18=1.56…z=.0594
.0594.0202=0.0392…=39 which is a C
(1D) P(707067/18=0.17…Z=.5675
7967/18=0.67…Z=.7486
.7486.5675=0.1811…Jill scored below 18% value 1.8=96 which she would’ve got.May 5, 2014 at 12:32 am #17955
Chana Max, RNParticipantFor part b) i am not sure why you are getting 74.02. According to my calculation with z= 0.38 using the formula x=(standard deviation)(z) + mean so (18) (0.38) +67 =60.16 which would be the letter grade B
May 5, 2014 at 1:21 am #17956
Ezra HalleckParticipantThere seems to be general confusion on how the 3 representations fit together. There is the score (out of 100 say), there is the percentile (probability) and there is the grade. The grade is determined only by the percentile.
A. What Aziza did is correct.
B. As a letter grade, you need to do no work as what is given is the percentile, so that Sally’s grade is C since 35 is between 30 and 39. As a score, what Vivian did is correct since percentile corresponds directly what the normal distribution table gives you in the center. So Sally’s score was 60 (closest integer).
C. 30 percentile is the probability (.3), so going backward in the chart, we get a zscore of 0.52, and destandardizing gives (18) (0.52) +67= 58 (to nearest integer).
D. 80 percentile is the probability (.8), so going backward in the chart, we get a zscore of 0.84, and destandardizing gives (18) (0.84) +67= 82 (to nearest integer).
Finally, it is very important to write a sentence answering the question corresponding to each part. Perhaps, a student can do so to gain participation points.May 7, 2014 at 2:42 am #17968
New York ZhangMemberHere is my standard deviation graph. I didn’t know if I had to provide seperate graphs; so I only provide the general graph for this question.

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