Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the medianDistance(in m)0 - 1 1 - 2 2 - 3 3 - 4 4 - 5Number of Students 40 80 62 38 30
-
Subject:
Math -
Author:
quintonpatton -
Created:
1 year ago
Answers 2
Answer:
1.3553
Step-by-step explanation:
356&$6 the 35-@4-&35&45
25-#36+#37+37623
36-#46&$
ans1.3553
-
Author:
marguerite3lus
-
Rate an answer:
1
The median distance is 2.08m
Step-by-step explanation:
Given:N=250
Distance (in m) = 0-1 1-2 2-3 3-4 4-5
Number of students= 40 80 62 38 30
To find:Median distance
Solution:First, we calculate the less than cumulative frequency using the number of students or frequencies given-
The c.f. for the first class interval 0-1 will be the same as the frequency=40
The c.f. for the class interval 1-2 will be 40+80= 120
c.f. for the class interval 2-3 = 120+62= 182
c.f. for the class interval 3-4 = 182+38 = 220
c.f. for the class interval 4-5 = 220+30 = 250
N=250, [tex]\frac{N}{2}[/tex]= [tex]\frac{250}{2}[/tex]
∴ [tex]\frac{N}{2}[/tex]= 125th term
The closest and greater than c.f. to 125 is 182
Hence, the corresponding class 2-3 is the median class
Now, Median= [tex]l+\frac{\frac{N}{2}-c.f.}{f} h[/tex]
where [tex]l[/tex]= lower limit of the median class = 2
[tex]\frac{N}{2}[/tex]=125
[tex]c.f.[/tex]= cumulative frequency of the class preceding median class=120
[tex]f[/tex]= frequency of the median class=62
[tex]h[/tex]= difference between the upper and lower limit of the median class
= 3-2 = 1
Median= [tex]l+\frac{\frac{N}{2}-c.f.}{f} h[/tex]
=[tex]2+\frac{125-120}{62} 1[/tex]
=[tex]2+\frac{5}{62} (1)[/tex]
= [tex]2+\frac{5}{62}[/tex]
∴ Median= [tex]2+0.0806[/tex]
∴ Median = 2.08m
⇒ The median distance jumped by the students is 2.08m
-
Author:
skipperbranch
-
Rate an answer:
4