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NCERT solutions for Class 11 Maths chapter 15 miscellaneous exercise has questions mainly on the concepts which have been discussed in the previous exercises like mean and standard deviation, variability and consistency etc. NCERT book Class 11 Maths chapter 15 miscellaneous exercise provides questions which can be helpful for the revision purpose as well. Class 11 Maths chapter 15 miscellaneous exercise is a good source to enhance one's marks in Boards as well as competitive examinations.
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Similarly, Many concepts will be used simultaneously in a question which can help recall previously read exercises. Below provided is the Class 11 Maths chapter 15 miscellaneous exercise solutions with other exercises as well. Students can refer to these for more practice in this chapter.
Answer:
Given,
The mean and variance of 8 observations are 9 and 9.25, respectively
Let the remaining two observations be x and y,
Observations: 6, 7, 10, 12, 12, 13, x, y.
∴ Mean,
60 + x + y = 72
x + y = 12 -(i)
Now, Variance
(Using (i))
-(ii)
Squaring (i), we get
(iii)
(iii) - (ii) :
2xy = 64 (iv)
Now, (ii) - (iv):
-(v)
Hence, From (i) and (v):
x – y = 4 x = 8 and y = 4
x – y = -4 x = 4 and y = 8
Therefore, The remaining observations are 4 and 8. (in no order)
Answer:
Given,
The mean and variance of 7 observations are 8 and 16, respectively
Let the remaining two observations be x and y,
Observations: 2, 4, 10, 12, 14, x, y
∴ Mean,
42 + x + y = 56
x + y = 14 -(i)
Now, Variance
(Using (i))
-(ii)
Squaring (i), we get
(iii)
(iii) - (ii) :
2xy = 96 (iv)
Now, (ii) - (iv):
-(v)
Hence, From (i) and (v):
x – y = 2 x = 8 and y = 6
x – y = -2 x = 6 and y = 8
Therefore, The remaining observations are 6 and 8. (in no order)
Answer:
Given,
Mean = 8 and Standard deviation = 4
Let the observations be
Mean,
Now, Let be the the resulting observations if each observation is multiplied by 3:
New mean,
= 24
We know that,
Standard Deviation =
-(i)
Now, Substituting the values of in (i):
Hence, the variance of the new observations =
Therefore, Standard Deviation = = = 12
Answer:
Given, Mean = and variance =
Now, Let be the the resulting observations if each observation is multiplied by a:
Hence the mean of the new observations is
We know,
Now, Substituting the values of :
Hence the variance of the new observations is
Hence proved.
If wrong item is omitted.
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 = 192
Therefore, Correct Mean = (Correct Sum)/19
= 10.1
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum - (8x8)
= 2080 – 64
= 2016
Hence, Correct Standard Deviation =
=2.02
If it is replaced by
Answer:
Given,
Number of observations, n = 20
Also, Incorrect mean = 10
And, Incorrect standard deviation = 2
Thus, incorrect sum = 200
Hence, correct sum of observations = 200 – 8 + 12 = 204
Therefore, Correct Mean = (Correct Sum)/20
= 10.2
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old sum - (8x8) + (12x12)
= 2080 – 64 + 144
= 2160
Hence, Correct Standard Deviation =
= 1.98
Which of the three subjects shows highest variability in marks and which shows the lowest?
Answer:
Given,
Standard deviation of physics = 15
Standard deviation of chemistry = 20
Standard deviation of mathematics = 12
We know ,
C.V.
The subject with greater C.V will be more variable than others.
Hence, Mathematics has lowest variability and Chemistry has highest variability.
Answer:
Given,
Initial Number of observations, n = 100
Thus, incorrect sum = 2000
Hence, New sum of observations = 2000 - 21-21-18 = 1940
New number of observation, n' = 100-3 =97
Therefore, New Mean = New Sum)/100
= 20
Now, Standard Deviation,
,which is the incorrect sum.
Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)
= 40900 - 441 - 441 - 324
= 39694
Hence, Correct Standard Deviation =
= 3.036
Miscellaneous exercise chapter 15 Class 11 chapter Statistics Is a good source for revision purposes specially and to check one’s level of understanding of this chapter as most of the questions are using multiple concepts. Class 11 Maths chapter 15 miscellaneous solutions mainly has 7 questions and it can take around 4-5 hours to complete the exercise. Hence, NCERT solutions for Class 11 Maths chapter 15 miscellaneous exercise is one stop solution for scoring better marks in the examination.
The Class 11 Maths chapter 15 exercise has a set of good questions which must be practiced before the examination.
Class 11 Maths chapter 15 miscellaneous exercises provides holistic understanding of the whole chapter as most of the concepts are used in this exercise.
These Class 11 Maths chapter 15 miscellaneous exercises is very rewarding from marks point of view in the examination.
Also see-
There are total 7 questions and some questions have subparts as well
Statistics can be used to analyse the given data in order to understand it in a better way.
Yes, Sometimes in the Board exam the questions are repeated.
Moderate to difficult level of questions are provided in this chapter.
No, as it has some good questions, miscellaneous exercise must be done.. It also helps in revising the earlier concepts.
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