Statistics Class 11 OP Malhotra Exe-28A ISC Maths Ch-28 Solutions. In this article you would learn about Combined Standard Deviations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Statistics Class 11 OP Malhotra Exe-28A ISC Maths Solutions Ch-28
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-28 | Statistics |
| Writer | O.P. Malhotra |
| Exe-28(A) | Combined Standard Deviations |
Combined Standard Deviations
Statistics Class 11 OP Malhotra Exe-28A ISC Maths Ch-28 Solutions.
Very Short Answer Type Questions.
Que-1: One set of 100 observations has the mean 15 and another set of 150 observations has the mean 16. Find the mean of 250 observations by combining the two sets of given observations.
Sol: Here, n1 = 100 ; x¯1 = 15 and n2 – 150 ; x¯2 = 16
∴ combined Mean x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
= (100×15+150×16)/(100+150)
= (1500+2400)/250
= 3900/250 = 15.6
Que-2: The mean age of 40 students is 6 years and the mean age of another group of 60 students is 20 years. Find out the mean age of the 100 students combined together.
Sol: Here n1 = 40; x¯1 = 16 and n2 = 60; x¯2 = 20
∴ combined Mean x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
= (40×16+60×20)/(40+60)
= (640+1200)/100
= 1840/100 = 18.4
Thus required mean age be 18.4 years.
Que-3: The mean of marks obtained in an examination by a group of 100 students is found to be 49.46. The mean of the marks obtained in the same examination by another group of 200 students was 52.32. Find the mean of the marks obtained by both the groups of students taken together.
Sol: Here, n1 = 100; x¯1 = 49. 46 and n2 = 200; x¯2 = 52.32
Thus marks obtained by first group of students = n1 x¯1
and marks obtained by second group of students = n2 x¯2
Hence, required mean marks obtained by both groups of students taken together = (n1x¯1+n2x¯2)/(n1+n2)
= (100×49.46+200×52.32)/(100+200)
= (4946+10464)/300 = 51.366
Que-4: The number of students in section X A and X B are 30 and 35 respectively. The mean scores of students in the mathematics test are as follows:
Find the mean score of XB.
Sol: Given n1 = 30; n2 = 35; x¯1 = 70; x¯12 = 62; x¯2 = ?
We know that x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
⇒ 62 = (30×70+35×x¯2)/(30+35)
⇒ 4030 – 2100 = 35x¯2
⇒ x¯2 = 1930/35
= 55.14
Que-5: Two samples of sizes 50 and 100 are given. The mean of these samples respectively are 56 and 50. Find the mean of size 150 by combining.
Sol: Here n1 = 50; n2 = 100; x¯1 = 56; x¯2 = 50
Then x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
⇒ x¯12 = (50×56+100×50)/(50+100)
⇒ x¯12 = (2800+5000)/150 = 7800/150 = 52
Que-6: The mean and standard deviation of the marks obtained by the groups of students, consisting of 50 each, are given below. Calculate the mean and standard deviation of the marks obtained by all the 100 students.
Sol: Here n1 = n2 = 50; x¯1 = 60; x¯2 = 55; σ1 = 8; σ1 = 7
Then combined mean x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
= (50×30+50×44)/(50+50) = (3000+2750)/100
= 5750/100 = 57.5
∴ d1 = x¯12 – x¯1 = 57.5 – 60 = – 2.5 and d2 = x¯12 – x¯2 = 57.5 – 55 = 2.5
Therefore, combined S.D = σ12 =
Que-7: The mean and standard deviation of distribution of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Sol: Here n1 = 100; n2 = 150; x¯1 = 50; x¯2 = 40
σ1 = 5; σ2 = 6
∴ Combined Mean x¯12 = (n1x¯1+n2x¯2)/(n1+n2)
= (100×50+150×40)/(100+150)
= 11000/250 = 44
∴ d1 = x¯12 – x¯1 = 44 – 50 = – 6 and d2 = x¯12 – x¯2 = 44 – 40 = 4
–: End of Statistics Class 11 OP Malhotra Exe-28A ISC Maths Ch-28 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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