Question

ine students in Mr. Ruiz's class all took a spelling test. Raw scores can range between 0 (0%) and 10 (100%). Their scores are listed below.

Student	Raw Score
A	4
B	5
C	*5*
D	6
E	6
F	6
G	7
H	7
I	8

The class mean is 6, median is 6 , mode is 6 , range is 4 , and standard deviation is 1.1(round to one decimal place).

Look at Student C's score. The score is in the 33 th percentile rank (round to the nearest integer). Its Z score is -0.90 (round to two decimal places, e.g., if you got 0.553, enter .55 or 0.55). Its T score is 41 (round to one decimal place, e.g., if you got 67.38, enter 67.4).

Nine students in Mr. Sotelo's class all took a spelling test. Raw scores can range between 0 (0%) and 10 (100%). Their scores are listed below.

Student	Raw Score
J	2
K	3
L	4
M	4
N	*5*
O	6
P	10
Q	10
R	10

The class mean is 6 , median is 5 , mode 10 is , range is 8 , and standard deviation is 3.2 (round to one decimal place).

Look at Student N's score. The score is in the 50 th percentile rank (round to the nearest integer). Its Z score is -0.33(round to two decimal places). Its T score is -0.9 (round to one decimal place).

What if a 80% correct score was used as a criterion to determine success ? What proportion of students in each class would meet that criterion?

Mr. Ruiz's class:________ % (Round to the nearest integer)

Mr. Sotelo's class:_______ % (Round to the nearest integer)

Answer 1

Solution

Mr. Ruiz's class: mean, µ = 6, standard deviation, σ = 1.1.

80% => raw score = 8.

Z-score for 8 = (8 - 6)/1.1 = 1.8182

P(Z > 1.8182) = 0.0345 ie success percentage = 3.45% or 3% Answer 1

Mr. Sotelo's class: mean, µ = 6, standard deviation, σ = 3.2.

80% => raw score = 8.

Z-score for 8 = (8 - 6)/3.2 = 0.625

P(Z > 0.625) = 0.266 ie success percentage = 26.6% or 27% Answer 2

DONE