find 1/2 of 600 kg fish
Answers
Answer:
Question 1: If weight classes (in kgs) are 50-52, 52-54, 54-56, 56-58, 58-60 and the corresponding students are 17, 35, 28, 15, 5. Which of the following options depict the class mark?
(a) 51, 53, 55, 57, 59
(b) 52, 54, 56, 58, 60
(c) 51, 55, 53, 57, 59
Solution 1:
We have been given the classes and the respective frequencies, and we're asked to find out the class mark.
What is the class mark?
Class mark is given by the sum of the lower limit and the upper limit of a particular class divided by 2. It is known as the mid-value of a given class.
\boxed{\sf Class \ mark = \dfrac{Lower \ limit + Upper \ Limit}{2}}
Class mark=
2
Lower limit+Upper Limit
The lower limit refers to the lowest value of the class, whereas the upper limit refers to the highest limit of the same class.
Using the formula mentioned above we get,
\begin{gathered}\left|\begin{array}{c | c | c}\sf \ Class \ & \sf Frequency & \sf Class mark \\\sf \ 50 - 52 \ & \sf 17 & \sf (50 + 52)/2 = 102/2 = \textsf{\textbf{51}} \\\sf \ 52 - 54 \ & \sf 35 & \sf (52 + 54)/2 = 106/2 = \textsf{\textbf{53}} \\\sf \ 54 - 56 \ & \sf 28 & \sf (54 + 56)/2 = 110/2 = \textsf{\textbf{55}} \\\sf \ 56 - 58 \ & \sf 15 & \sf (56 + 58)/2 = 114/2 = \textsf{\textbf{57}} \\\sf \ 58 - 60 \ & \sf 5 & \sf (58 + 60)/2 = 118/2 = \textsf{\textbf{59}} \\\end{array}\right|\end{gathered}
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Class
50−52
52−54
54−56
56−58
58−60
Frequency
17
35
28
15
5
Classmark
(50+52)/2=102/2=51
(52+54)/2=106/2=53
(54+56)/2=110/2=55
(56+58)/2=114/2=57
(58+60)/2=118/2=59
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Therefore, the correct option is (a) 51, 53, 55, 57, 59.
Question 2: In the above question, if we take the assumed mean to be 55 and h = 2, what is the value of ∑fu?
(a) 44
(b) -54
(c) -44
Solution 2:
According to the question,
Assumed mean (a) = 55
Height of the class (h) = 2
∑fu
\begin{gathered}\left|\begin{array}{c | c | c | c }\sf \ Class \ & \sf Frequency \ (f) & \sf Class mark \ (x)& \sf u = (x - a)/h \\\sf \ 50 - 52 \ & \sf 17 & \sf 51 & \sf (51 - 55)/2 = \textbf{-2}\\\sf \ 52 - 54 \ & \sf 35 & \sf 53 & \sf (53 - 55)/2 = \textbf{-1} \\\sf \ 54 - 56 \ & \sf 28 & \sf 55 = a & \sf (55 - 55)/2 = \textbf{0} \\\sf \ 56 - 58 \ & \sf 15 & \sf 57 & \sf (57 - 55)/2 = \textbf{1} \\\sf \ 58 - 60 \ & \sf 5 & \sf 59 & \sf (59 - 55)/2 = \textbf{2} \\\end{array}\right|\end{gathered}
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Class
50−52
52−54
54−56
56−58
58−60
Frequency (f)
17
35
28
15
5
Classmark (x)
51
53
55=a
57
59
u=(x−a)/h
(51−55)/2=-2
(53−55)/2=-1
(55−55)/2=0
(57−55)/2=1
(59−55)/2=2
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We've got the value of "u" now, so let's find out the value of fu for each class,
\begin{gathered}\left|\begin{array}{c | c | c | c | c }\sf \ Class \ & \sf Frequency \ (f) & \sf Class mark \ (x)& \sf u = (x - a)/h & \sf \ Fu \ \\\sf \ 50 - 52 \ & \sf 17 & \sf 51 & \sf -2 & \sf 17 \times -2 = -34 \\\sf \ 52 - 54 \ & \sf 35 & \sf 53 & \sf -1 & \sf 35 \times -1 = -35 \\\sf \ 54 - 56 \ & \sf 28 & \sf 55 = a & \sf 0 & \sf 0 \\\sf \ 56 - 58 \ & \sf 15 & \sf 57 & \sf 1 & \sf 15 \times 1 = 15 \\\sf \ 58 - 60 \ & \sf 5 & \sf 59 & \sf 2 & \sf 5 \times 2 = 10 \\\end{array}\right|\end{gathered}
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Class
50−52
52−54
54−56
56−58
58−60
Frequency (f)
17
35
28
15
5
Classmark (x)
51
53
55=a
57
59
u=(x−a)/h
−2
−1
0
1
2
Fu
17×−2=−34
35×−1=−35
0
15×1=15
5×2=10
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∑fu = (-34) + (-35) + 0 + 15 + 10
∑fu = -69 + 25
∑fu = -44
Therefore, the correct option is (c) -44
Question 3: What is the value of mean weight for the above question?
(a) 55 kg
(b) -0.44 kg
(c) 54.46 kg
(d) 54.12 kg
\sf \Longrightarrow Mean = a + \dfrac{\Sigma fu}{\Sigma f}⟹Mean=a+
Σf
Σfu
\sf \Longrightarrow Mean = 55 + \dfrac{-44}{17 + 35 + 28 + 15 + 5}⟹Mean=55+
17+35+28+15+5
−44
\sf \Longrightarrow Mean = 55 + \dfrac{-44}{100}⟹Mean=55+
100
−44
\sf \Longrightarrow Mean = 55 - 0.44⟹Mean=55−0.44
\sf \Longrightarrow \textsf{\textbf{Mean = 54.56}}⟹Mean = 54.56
Therefore, the correct option is (c) 54.56
Answer:
300 kg fish
Step-by-step explanation:
1/2×600
1×300
300 ANS