Please answer the first part of qn 30....
Answers
Answer:
x = 8, y = 7
Step-by-step explanation:
Class Frequency c.f
0-10 5 5
10-20 x 5 + x
20-30 20 20 + 5 + x = 25 + x
30-40 15 25 + 15 + x = 40 + x
40-50 y 40 + x + y
50-60 5 45 + x + y
Total 60
Now,
From the table it can be observer that n = 60.
45 + x + y = 60
x + y = 15 ----- (i)
Given:
Median = 28.5.
Range = Median class = 20 - 30.
Lower limit l = 20.
Cumulative Frequency(cf) = 5 + x.
Frequency of median class (f)= 20.
class size(h) = 10.
∴ Median = l + [n/2 - cf/f] * h
⇒ 28.5 = 20 + [30 - (5 + x)/20] * 10
⇒ 28.5 = 20 + [25 - x/20] * 10
⇒ 28.5 = 20 + [25 - x]/2
⇒ 57 = 40 + 25 - x
⇒ 57 - 65 = -x
⇒ x = 8
Substitute x = 8 in (i), we get
⇒ x + y = 15
⇒ 8 + y = 15
⇒ y = 7
Therefore, the value of x = 8 and y = 7.
Hope it helps!
Answer:
ıllıllı hey ıllıllı
Step-by-step explanation:
Median:
Median is that value of the given observation which divides it into exactly two parts.
MEDIAN for the GROUPED data :
For this we find the Cumulative frequency(cf) of all the classes and n/2 , where n = number of observations.
Now find the class whose Cumulative frequency is greater than and nearest to n/2 and this class is called median class,then use the following formula calculating the median.
MEDIAN = l + [(n/2 - cf )/f ] ×h
Where,
l = lower limit of the median class
n = number of observations
cf = cumulative frequency of class interval preceed the median class
f = frequency of median class
h = class size
n = 60 and hence n /2 = 30
Median class is 20 – 30 with cumulative frequency = 25 + x
lower limit of median class = 20, c f = 5 + x , f = 20 and h = 10
Median = l + n /2 − c f / f × h
Or,
28.5 = 20 + 30 − 5 − x / 20 × 10
Or, 25 − x / 2 = 8.5
Or, 25 − x = 17
Or, x = 25 − 17 = 18
Now, from cumulative frequency, we can find the value of x + y as follows:
60 = 5 + 20 + 15 + 5 + x + y
Or, 45 + x + y = 60
Or, x + y = 60 − 45 = 15
Hence,
y = 15 − x = 15 − 8 = 7
Hence, x = 8 and y = 7