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Chapter 2: Relations and Functions
Chapter 3: Trigonometric Functions
Chapter 4: Principle of Mathematical Induction
Chapter 5: Complex Numbers and Quadratic Equations
Chapter 6: Linear Inequalities
Chapter 7: Permutations and Combinations
Chapter 8: Binomial Theorem
Chapter 9: Sequences and Series
Chapter 10: Straight Lines
Chapter 11: Conic Sections
Chapter 12: Introduction to Three Dimensional Geometry
Chapter 13: Limits and Derivatives
Chapter 14: Mathematical Reasoning
Chapter 15: Statistics
Chapter 16: Probability
Chapter 15: Statistics
NCERT solutions for Mathematics Exemplar Class 11 Chapter 15 Statistics Solved Examples [Pages 272 - 277]
Find the mean deviation about the mean of the following data:
Find the variance and standard deviation for the following data: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Show that the two formulae for the standard deviation of ungrouped data.
`sigma = sqrt((x_i - barx)^2/n)` and `sigma`' = `sqrt((x^2_i)/n - barx^2)` are equivalent.
Calculate variance of the following data:
|4 – 8||3|
|8 – 12||6|
|12 – 16||4|
|16 – 20||7|
Mean `(barx) = (f_ix_i)/(f_i) = (3 xx 6 + 6 xx 10 + 4 xx 14 + 7 xx 18)/20` = 13
Calculate mean, variation and standard deviation of the following frequency distribution:
|1 – 10||11|
|10 – 20||29|
|20 – 30||18|
|30 – 40||4|
|40 – 50||5|
|50 – 60||3|
Life of bulbs produced by two factories A and B are given below:
|Length of life
(Number of bulbs)
(Number of bulbs)
|550 – 650||10||8|
|650 – 750||22||60|
|750 – 850||52||24|
|850 – 950||20||16|
|950 – 1050||16||12|
The bulbs of which factory are more consistent from the point of view of length of life?
Objective Type Questions from 7 to 9
The mean deviation of the data 2, 9, 9, 3, 6, 9, 4 from the mean is ______.
Variance of the data 2, 4, 5, 6, 8, 17 is 23.33. Then variance of 4, 8, 10, 12, 16, 34 will be ______.
A set of n values x1, x2, ..., xn has standard deviation 6. The standard deviation of n values x1 + k, x2 + k, ..., xn + k will be ______.
σ + k
σ – k
NCERT solutions for Mathematics Exemplar Class 11 Chapter 15 Statistics Exercise [Pages 278 - 238]
Find the mean deviation about the mean of the distribution:
Find the mean deviation about the median of the following distribution:
|No. of students||2||3||8||3||4|
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an odd number.
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
Find the standard deviation of the first n natural numbers.
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:
Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations x1, x2, ..., x15, also in seconds, is now available and we have `sum_(i = 1)^15 x_i` = 279 and `sum_(i = 1)^15 x^2` = 5524. Calculate the standard derivation based on all 40 observations.
The mean and standard deviation of a set of n1 observations are `barx_1` and s1, respectively while the mean and standard deviation of another set of n2 observations are `barx_2` and s2, respectively. Show that the standard deviation of the combined set of (n1 + n2) observations is given by
S.D. = `sqrt((n_1(s_1)^2 + n_2(s_2)^2)/(n_1 + n_2) + (n_1n_2 (barx_1 - barx_2)^2)/(n_1 + n_2)^2)`
Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.
The frequency distribution:
where A is a positive integer, has a variance of 160. Determine the value of A.
For the frequency distribution:
Find the standard deviation.
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test.
|Frequency||x – 2||x||x2||(x + 1)2||2x||x + 1|
where x is a positive integer. Determine the mean and standard deviation of the marks.
The mean life of a sample of 60 bulbs was 650 hours and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and standard deviation 7 hours. Find the overall standard deviation.
Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.
If for distribution `sum(x - 5)` = 3, `sum(x - 5)^2` = 43 and total number of items is 18. Find the mean and standard deviation.
Find the mean and variance of the frequency distribution given below:
|`x`||1 ≤ x < 3||3 ≤ x < 5||5 ≤ x < 7||7 ≤ x < 10|
Calculate the mean deviation about the mean for the following frequency distribution:
|Class interval||0 – 4||4 – 8||8 – 12||12 – 16||16 – 20|
Calculate the mean deviation from the median of the following data:
|Class interval||0 – 6||6 – 12||12 – 18||18 – 24||24 – 30|
Determine the mean and standard deviation for the following distribution:
The weights of coffee in 70 jars is shown in the following table:
|200 – 201||13|
|201 – 202||27|
|202 – 203||18|
|203 – 204||10|
|204 – 205||1|
|205 – 206||1|
Determine variance and standard deviation of the above distribution.
Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
Who is more intelligent and who is more consistent?
Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Objective Type Questions from 24 to 39
The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is ______.
Mean deviation for n observations x1, x2, ..., xn from their mean `barx` is given by ______.
`sum_(i = 1)^n (x_i - barx)`
`1/n sum_(i = 1)^n |x_i - barx|`
`sum_(i = 1)^n (x_i - barx)^2`
`1/n sum_(i = 1)^n (x_i - barx)^2`
When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is ______.
Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is ______.
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is ______.
Let x1, x2, ..., xn be n observations and `barx` be their arithmetic mean. The formula for the standard deviation is given by ______.
(x_i - barx)^2`
`(x_i - barx)^2/n`
`sqrt((x_i - barx)^2/n`
`sqrt(x^2/n + barx^2)`
The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is ______.
Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is ______.
s + k
Let x1, x2, x3, x4, x5 be the observations with mean m and standard deviation s. The standard deviation of the observations kx1, kx2, kx3, kx4, kx5 is ______.
k + s
Let x1, x2, ... xn be n observations. Let wi = lxi + k for i = 1, 2, ...n, where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of wi’s is 55 and standard deviation of wi’s is 15, the values of l and k should be ______.
l = 1.25, k = – 5
l = – 1.25, k = 5
l = 2.5, k = – 5
l = 2.5, k = 5
Standard deviations for first 10 natural numbers is ______.
Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is ______.
Consider the first 10 positive integers. If we multiply each number by –1 and then add 1 to each number, the variance of the numbers so obtained is ______.
The following information relates to a sample of size 60 `sumx^2` = 18000 and `sumx` = 960, then the variance is ______.
Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is ______.
The standard deviation of some temperature data in °C is 5. If the data were converted into ºF, the variance would be ______.
Fill in the blanks 40 to 46
Coefficient of variation = `.../"Mean" xx 100`
If `barx` is the mean of n values of x, then `sum_(i = 1)^n (x_i - barx)` is always equal to ______. If a has any value other than `barx`, then `sum_(i = 1)^n (x_i - barx)^2` is ______ than `sum(x_i - a)^2`
If the variance of a data is 121, then the standard deviation of the data is ______.
The standard deviation of a data is ______ of any change in orgin, but is ______ on the change of scale.
The sum of squares of the deviations of the values of the variable is ______ when taken about their arithmetic mean.
The mean deviation of the data is ______ when measured from the median.
The standard deviation is ______to the mean deviation taken from the arithmetic mean.
Chapter 15: Statistics
NCERT solutions for Mathematics Exemplar Class 11 chapter 15 - Statistics
NCERT solutions for Mathematics Exemplar Class 11 chapter 15 (Statistics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Exemplar Class 11 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Mathematics Exemplar Class 11 chapter 15 Statistics are Central Tendency - Median, Measures of Dispersion, Concept of Range, Mean Deviation, Introduction of Variance and Standard Deviation, Standard Deviation, Standard Deviation of a Discrete Frequency Distribution, Standard Deviation of a Continuous Frequency Distribution, Shortcut Method to Find Variance and Standard Deviation, Introduction of Analysis of Frequency Distributions, Comparison of Two Frequency Distributions with Same Mean, Statistics Concept, Central Tendency - Mean, Measures of Dispersion - Quartile Deviation, Standard Deviation - by Short Cut Method, Concept of Mode.
Using NCERT Class 11 solutions Statistics exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 11 prefer NCERT Textbook Solutions to score more in exam.
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