The mean of the following frequency distribution is 25. Find the missing frequency f: Class 0-10(5),10-20(8),20-30(12),30-40(f),40-50(3)
10
11
12
13
Explanation: Midpoints: $5, 15, 25, 35, 45$. $\sum f = 5+8+12+f+3 = 28+f$. $\sum fx = 5 \times 5 + 8 \times 15 + 12 \times 25 + 35f + 3 \times 45 = 580 + 35f$. Setting the mean to $25$: $\frac{580+35f}{28+f}=25 \Rightarrow 580+35f=700+25f \Rightarrow 10f=120 \Rightarrow f=12$.
Question 2 of 6medium
The median of the distribution in Question 1 (with f=12) is:
24.83
25.83
26.83
27.83
Explanation: $N=40$, so $\frac{N}{2}=20$. The cumulative frequencies are $5, 13, 25, 37, 40$, so the median class is $20\text{-}30$ with $l=20,\ cf=13,\ f=12,\ h=10$. Median $= 20 + \frac{20-13}{12} \times 10 = 20 + \frac{7}{12} \times 10 = 20 + 5.83 = 25.83$.
Question 3 of 6medium
If the mode of the following data is 36, find x: Class 0-20(6),20-40(10),40-60(x),60-80(8),80-100(4)
7
8
9
10
Explanation: Since the mode $36$ lies in $20\text{-}40$, that is the modal class with $l=20,\ f_1=10,\ f_0=6,\ f_2=x,\ h=20$. Mode $= l + \frac{f_1-f_0}{2f_1-f_0-f_2} \times h = 20 + \frac{10-6}{20-6-x} \times 20 = 36$. So $\frac{80}{14-x}=16 \Rightarrow 80=224-16x \Rightarrow 16x=144 \Rightarrow x=9$.
Question 4 of 6medium
The less than ogive and greater than ogive for a distribution intersect at (35,50). The median is:
35
50
85
25
Explanation: Intersection point's x-coordinate is the median = 35
Question 5 of 6medium
Find the mean of the following data using step deviation method with a=25, h=10: Class 0-10(4),10-20(6),20-30(10),30-40(8),40-50(2)
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