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The question is incomplete. The complete question is
A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.
Which class should get the reward, and why?
- The 2nd-period class should get the reward. They have the highest score, a perfect 100.
- The 2nd-period class should get the reward. They have a higher median.
- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.
- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.
The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is
In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.
Hence, the mean of the results of the 4th period is
Hence, obviously, the 4th period is having more mean.
Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.
So, just by observing the box plot, we can give the statement that "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "
Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.
Learn more about box plots here-
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Answer:
a) the K-map is in the attachment
<em>f = Σ</em>m(0,1,2,3,6,10,14,15)
b) from the k-map, the minimum sum of products is
F = A'B' + CD' + ABC
c) the minimum product of sums is
F = (B' + C)(A' + C)(A+ B' +D')(A' + B + D')
Step-by-step explanation:
A Karnaugh map (K-map) is a pictorial framework used to limit the Boolean expressions without utilizing Boolean algebra theorems and equation controls.
a) the given function is <em>f</em>(A,B,C,D)=A‘B’+CD’+ABC +A’B’CD’+ABCD’
expanding the function as four variable terms
<em>f</em>(A,B,C,D)=A‘B’+CD’+ABC +A’B’CD’+ABCD’
= A'B'(C + C')(D + D')+(A + A')(B + B")CD' + ABC(D + D') + A'B'CD' + ABCD'
= A'B'CD + A'B'CD' + A'B'C'D' + ABCD' +AB'CD' + A'BCD' + A'B'CD' + ABCD +ABCD' + A'B'CD' + ABCD'
=A'B'CD + A'B'CD' + A'B'C'D + A'B'C'D' + ABCD' + AB'CD' + A'BCD' +ABCD
<em>f = Σ</em>m(0,1,2,3,6,10,14,15)
note: diagram is in the attachment
b) the minterms for the minimum sum of product are
<em>f = Σ</em>m(0,1,2,3,6,10,14,15)
simplifying the K-map(done in the attachment)
from the k-map, the minimum sum of products is
F = A'B' + CD' + ABC
c) the maxterms for the minimum product of sums are
<em>f </em>= ПM(4,5,7,8,9,11,12,13)
plot the K-map to find minimum product of sums(done in the attachment)
the minimum product of sums is
F = (B' + C)(A' + C)(A+ B' +D')(A' + B + D')
