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Assessment Task For Lesson 3.2: X X W W W W X X

The document contains 4 assessment problems with solutions: 1) Calculating the weighted mean price of colas sold in different sizes. The weighted mean price was $35.64. 2) Determining a student's test score based on a normal distribution with mean 76 and standard deviation 8. The student's score was 71. 3) Comparing a student's performance in Math and Science based on their z-scores. The student performed better in Science. 4) Constructing a boxplot to describe a data set ranging from 45 to 59 with a median of 53.

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Cristina Roche
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2K views2 pages

Assessment Task For Lesson 3.2: X X W W W W X X

The document contains 4 assessment problems with solutions: 1) Calculating the weighted mean price of colas sold in different sizes. The weighted mean price was $35.64. 2) Determining a student's test score based on a normal distribution with mean 76 and standard deviation 8. The student's score was 71. 3) Comparing a student's performance in Math and Science based on their z-scores. The student performed better in Science. 4) Constructing a boxplot to describe a data set ranging from 45 to 59 with a median of 53.

Uploaded by

Cristina Roche
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Tuesday, 1:30 – 3:30 PM

ASSESSMENT TASK FOR LESSON 3.2


Answer the following problems:
1. A pizza parlor sells colas in three sizes: small, medium, and large. The small size costs
P25, the medium P35, and the large P50. Yesterday, 120 small, 250 medium, and 100
large colas were sold. What was the weighted mean price per cola?

x 1 w 2+ x 2 w2 + x 3 w 3
x w=
w1 + w2 +w3
25(120)+35(250)+50(100)
x w=
120+250+100
x w =35.63829787 ≈ 35.64
Therefore, the weighted mean price per cola is P35.64.

2. A class of 300 students had test scores that are adequately described by a normal
distribution with a mean of 76 and a standard deviation of 8. If a certain student had a z-
score of -0.625, determine his/her test scores.
x−x
z=
s
x=zs+ x
x=(−0.625)(8)+ 76
x=71
Therefore, the certain student had a score of 71.
3. The report card of Alvin shows that his grade in Math is 89 and Science is 93. The mean
grade in Math is 85, and the standard deviation is 5. In Science, the mean grade is 80, and
the standard deviation is 8. In which subject does Alvin perform better?

x−x 89−85
In Math, z= = =¿ 0.8
s 5
x−x 93−80
In Science, z= = =¿ 1.625
s 8
The values show that the Math is 0.8 higher than the mean, while Science is 1.625
standard deviation higher than the mean. Therefore, Alvin performs better in Science.
4. Construct a boxplot for the data set: 53, 45, 59, 48, 54, 46, 51, 58, 55. Describe the
distribution of the data set.

45, 46, 48, 51, 53, 54, 55, 58, 59


Median: 53
Lower Extreme: 45
Higher Extreme: 59
These are the endpoints of our whiskers.

45 46 48 51 53 54 55 58 59
Lower Extreme Median Higher Extreme

Average of these two boxed Average of these two boxed


numbers to get the end of the first numbers to get the beginning of the
quartile. last quartile.
46+ 48 55+58
=47 =56.5
2 2
Graph.

25 30 35 40 45 50 55 60 65

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