Statistics

Given: Data distribution

To find: the mean deviation about the mean of the distribution

Let us make a table of the given data and append other columns after calculations

Here mean,

So the above table with more columns is as shown below,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the mean of the distribution is 1.25

Given: Data distribution

To find: the mean deviation about the median

Let us make a table of the given data and append other columns after calculations

Now, here N=20, which is even.

Here median,

Both these observations lie in cumulative frequency 13, for which corresponding observation is 12.

So the above table with more columns is as shown below,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the median of the distribution is 1.25

Given: set of first n natural numbers when n is an odd number

To find: the mean deviation about the mean

We know first n natural numbers are 1, 2, 3….., n. And given n is odd number.

So mean is,

The deviations of numbers from the mean are as shown below,

Or,

Or,

So the absolute values of deviation from the mean is

The sum of absolute values of deviations from the mean, is

i.e., 2 times sum of terms, so it can be written as

Therefore, mean deviation about the mean is

Hence the mean deviation about the mean of the set of first n natural numbers when n is an odd number is

Given: set of first n natural numbers when n is an even number.

To find: the mean deviation about the mean

We know first n natural numbers are 1, 2, 3….., n. And given n is even number.

So mean is,

The deviations of numbers from the mean are as shown below,

Or,

Or,

So the absolute values of deviation from the mean is

The sum of absolute values of deviations from the mean, is

Now we know sum of first n natural numbers = n²

Therefore, mean deviation about the mean is

Hence the mean deviation about the mean of the set of first n natural numbers when n is an even number is

Given: set of first n natural numbers

To find: the standard deviation

given first n natural numbers, we can write in table as shown below

So, the sums of these are

Therefore, the standard deviation can be written as,

Hence the standard deviation of the first n natural numbers is

Given: Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Another set of 15 observations x₁, x₂, ..., x₁₅, also in seconds, is and

To find: the standard derivation based on all 40 observations

As per the given criteria,

In first set,

Number of observations, n₁=25

Mean,

And standard deviation,

And

In second set,

Number of observations, n₂=15

and

For the first set we have

∑x_i=25×18.2=455

Therefore the standard deviation becomes,

Substituting the values, we get

For the combined standard deviation of the 40 observation, n=40

And

Therefore the standard deviation can be written as,

Substituting the values, we get

Therefore the standard deviation can be written as,

σ=3.87

Hence the standard derivation based on all 40 observations is 3.87.

Given: The mean and standard deviation of a set of n₁ observations are and s₁, respectively while the mean and standard deviation of another set of n₂ observations are and s₂, respectively

To show: the standard deviation of the combined set of (n₁ + n₂) observations is given by

As per given criteria,

For first set

Let x_i where i=1, 2, 3,4 , …, n₁

For second set

And y_j where j=1, 2, 3, 4, …, n₂

And the means are

Now mean of the combined series is given by

And the corresponding square of standard deviation is

Therefore, square of standard deviation becomes,

Now,

But the algebraic sum of the deviation of values of first series from their mean is zero.

Also,

But

Substituting value from equation (i), we get

Substituting this value in equation (iii), we get

Similarly, we have

But the algebraic sum of the deviation of values of second series from their mean is zero.

Also,

But

Substituting value from equation (i), we get

Substituting this value in equation (v), we get

Substituting equation (iv) and (vi) in equation (ii), we get

So the combined standard deviation

Hence proved

Given: 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.

To show: the standard deviation of the set obtained by combining the given two sets

As per given criteria,

For first set

Number of observations, n₁=20

Standard deviation, s₁=5

And mean,

For second set

Number of observations, n₂=20

Standard deviation, s₂=5

And mean,

We know the standard deviation for combined two series is

Substituting the corresponding values, we get

Or, σ=5.59

Hence the standard deviation of the set obtained by combining the given two sets is 5.59

Given: frequency distribution table, where variance =160

To find: the value of A, where A is a positive number

Let us make a table of the given data and append other columns after calculations

And we know variance is

Substituting values from above table and also given variance=160, we get

⇒ A²=49

⇒ A=7

Hence the value of A is 7.

Given: frequency distribution table

To find: the standard deviation

Let us make a table of the given data and append other columns after calculations

And we know standard deviation is

Substituting values from above table, we get

⇒ σ=1.37

Hence the standard deviation is 1.37.

Given: There are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

To find: the mean and standard deviation of the marks.

It is given there are 60 students in the class, so

∑f_i=60

⇒ (x-2)+x+x²+(x+1)²+2x+x+1=60

⇒ 5x-1 +x²+x²+2x+1=60

⇒ 2x² +7x=60

⇒ 2x² +7x-60=0

Splitting the middle term, we get

⇒ 2x² + 15x – 8x – 60 = 0

⇒ x(2x + 15) – 4(2x + 15) = 0

⇒ (2x + 15) (x – 4) = 0

⇒ 2x + 15 = 0 or x-4=0

⇒ 2x=-15 or x=4

Given x is a positive number, so x can take 4 as the only value.

And let assumed mean, a=3.

Now put x = 4 and a=3 in the frequency distribution table and add other columns after calculations, we get

And we know standard deviation is

Substituting values from above table, we get

⇒ σ=1.12

Hence the standard deviation is 1.12

Now mean is

=2.8

Hence the mean and standard deviation of the marks are 2.8 and 1.12 respectively.

Given: 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

To find: the overall standard deviation

As per given criteria,

In first set of samples,

Number of sample bulbs, n₁=60

Standard deviation, s₁=8hrs

Mean life,

And in second set of samples,

Number of sample bulbs, n₂=80

Standard deviation, s₂=7hrs

Mean life,

We know the standard deviation for combined two series is

Substituting the corresponding values, we get

Or, σ=8.9

Hence the standard deviation of the set obtained by combining the given two sets is 8.9

Given: Mean and standard deviation of 100 items are 50 and 4, respectively

To find: the sum of all items and the sum of the squares of the items

As per given criteria,

Number of items, n=100

Mean of the given items,

But we know,

Substituting the corresponding values, we get

⇒ ∑x_i=50× 100=5000

Hence the sum of all the 100 items = 5000

Also given the standard deviation of the 100 items is 4

i.e., σ=4

But we know

Substituting the corresponding values, we get

Now taking square on both sides, we get

And the sum of the squares of all the 100 items is 251600.

Given: for a distribution ∑ (x −5)=3, ∑ (x −5)² = 43 and the total number of item is 18

To find: the mean and standard deviation.

As per given criteria,

Number of items, n=18

And given ∑(x-5)=3,

And also given, ∑(x −5)² = 43

But we know mean can be written as,

Here assumed mean is 5, so substituting the corresponding values in above equation, we get

And we know the standard deviation can be written as,

Substituting the corresponding values, we get

Hence σ=1.54

So the mean and standard deviation of given items is 5.17 and 1.54 respectively.

Given: the frequency distribution

To find: the mean and variance

Converting the ranges of x to groups, the given table can be rewritten as shown below,

And we know variance can be written as

Substituting values from above table, we get

We also know mean can be written as

Substituting values from above table, we get

Hence the mean and variance of the given frequency distribution is 4.16 and 3.96 respectively.

Given: the frequency distribution

To find: the mean deviation about the mean

Let us make a table of the given data and append other columns after calculations

Here mean,

So the above table with more columns is as shown below,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the mean of the distribution is 3.84

Given: the frequency distribution

To find: the mean deviation from the median

Let us make a table of the given data and append other columns after calculations

Now, here N=20, which is even.

Here median class=,

This observation lie in the class interval 12-18, so median can be written as,

Here l=12, cf=9, f=3, h=6 and N=20, substituting these values, the above equation becomes,

⇒M=12+2=14

So the above table with more columns is as shown below,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the median of the distribution is 7

Given: the frequency distribution

To find: the mean and standard deviation

Let us make a table of the given data and append other columns after calculations

Here mean,

So the above table with more columns is as shown below,

And we know standard deviation is

Substituting values from above table, we get

⇒ σ=2.9

Hence the mean and standard deviation of the marks are 6 and 2.9 respectively.

Given: the weights of coffee in 70 jars

To find: the variance and standard deviation of the distribution

Let us make a table of the given data and append other columns after calculations

Here mean,

So the above table with more columns is as shown below,

And we know standard deviation is

Substituting values from above table, we get

⇒ σ=1.08g

And σ²=1.08²=1.17g

Hence the variance and standard deviation of the distribution are 1.166g and 1.08 respectively.

Given: first n terms of an A.P. whose first term is a and common difference is d

To find: mean and standard deviation

The given AP in tabular form is as shown below,

Here we have assumed a as mean.

Given the AP have n terms. And we know the sum of all the terms of AP can be written as,

Now we will calculate the actual mean,

Substituting the corresponding values, we get

Now we other two columns sums, i.e.,

Now we know standard deviation is given by,

Substituting the corresponding values, we get

Cancelling the like terms, we get

Taking out common terms we get

By taking the LCM, we get

Hence the mean and standard deviation of the given AP is and respectively.

Given: the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests

To find: who is more intelligent and who is more consistent

Case 1: For Ravi

The marks of Ravi being taken separately and finding other values can be tabulated as shown below,

Here we have assumed 45 as mean.

Total there are marks of 10 subjects.

Now we know standard deviation is given by,

Substituting the corresponding values, we get

σ=13

And mean is

Hence the mean and standard deviation of the Ravi is 44.1 and 13 respectively.

Case 1: For Hashina

The marks of Hashina being taken separately and finding other values can be tabulated as shown below,

Here as so 53 is mean.

Total there are marks of 10 subjects.

Now we know standard deviation is given by,

Substituting the corresponding values, we get

σ=24.35

Hence the mean and standard deviation of the Hashina is 53 and 24.35 respectively.

Now we will analyse them,

For Ravi,

For Hasina,

Now as CV(of Ravi)<CV of Hasina

Hence Ravi is more consistent.

Mean of Hasina>Mean of Ravi,

Hence Hasina is more intelligent.

Given: 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

To find: the correct standard deviation.

As per given criteria,

Number of observations, n=100

Mean of the given observations before correction,

But we know,

Substituting the corresponding values, we get

⇒ ∑x_i=40× 100=4000

It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively,

So ∑x_i=4000-30-70+3+27=3930

So the correct mean after correction is

Also given the standard deviation of the 100 observations is 10 before correction,

i.e., σ=10

But we know

Substituting the corresponding values, we get

Now taking square on both sides, we get

It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, so correction is

So the correct standard deviation after correction is

σ=10.24

Hence the corrected standard deviation is 10.24.

Given: 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

To find: the correct mean and the variance.

As per given criteria,

Number of reading, n=10

Mean of the given readings before correction,

But we know,

Substituting the corresponding values, we get

⇒ ∑x_i=45× 10=450

It is said one reading 25 was wrongly taken as 52,

So ∑x_i=450-52+25=423

So the correct mean after correction is

Also given the variance of the 10 readings is 16 before correction,

i.e., σ²=16

But we know

Substituting the corresponding values, we get

It is said one reading 25 was wrongly taken as 52, so

So the correct variance after correction is

σ²=1833.1-(42.3)²=1833.1-1789.29

σ²=43.81

Hence the corrected mean and variance is 42.3 and 43.81 respectively.

Given data is 3, 10, 10, 4, 7, 10, 5. They are total 7.

Here mean,

This can be written in table form as,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the mean of the distribution is 2.57.

We know for n observations x₁, x₂, ..., x_n having is given by

But we know

So mean deviation becomes,

Or

Given the lives (in hours) of 5 bulbs is 1357, 1090, 1666, 1494, 1623

Here mean,

This can be written in table form as,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the mean of the lives of 5 bulbs is 178

Given the marks obtained by 9 students in a mathematics test are 50, 69, 20, 33, 53, 39, 40, 65, 59

As number of students =9, which is odd.

So median will be = 5^th term.

Arranging these in ascending order, we get

20, 33, 39, 40, 50, 53, 59, 65, 69

So the 5^th term after arranging is 50,

So median is 50.

This can be written in table form as,

Hence Mean Deviation becomes,

Therefore, the mean deviation about the median of the marks of 9 subjects is 12.67

Given data in tabular form can be written as

But we know, standard deviation can be written as

Substituting the corresponding values, we get

Hence the standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is .

We know, standard deviation for x₁, x₂, ..., x_n observations can be written as

Where is the arithmetic mean

As per given criteria,

Number of observations, n=100

Mean of the given observations,

But we know,

Substituting the corresponding values, we get

⇒ ∑x_i=50× 100=5000

Hence the sum of all the 100 observations = 5000

Also given the standard deviation of the 100 observations is 5

i.e., σ=5

But we know

Substituting the corresponding values, we get

Now taking square on both sides, we get

And the sum of the squares of all the 100 observations is 252500.

Given observations are a, b, c, d, e

So the mean of the 5 observations is

⇒ ∑x_i=a+b+c+d+e=5m……..(i)

And the standard deviation of the 5 observations is

Substituting equation (i) in above equation we get

Now we will find the mean and standard deviation of the observations a + k, b + k, c + k, d + k, e + k, we get

So the mean of these 5 observations is

Substituting value from equation (i), we get

m₁=m+k…………(iii)

And the standard deviation of the 5 observations is

Substituting equation (iii) in above equation we get

Substituting the value from equation (i), we get

Cancelling like terms, we get

Comparing this with equation (ii), we get

σ=s

Hence the standard deviation of new set of observations, i.e., a + k, b + k, c + k, d + k, e + k is s, itself.

Given observations are x₁, x₂, x₃, x₄, x₅

So the mean of the 5 observations is

⇒ ∑x_i= x₁+x₂+ x₃+x₄+x₅ =5m……..(i)

And the standard deviation of the 5 observations is

Substituting equation (i) in above equation we get

Now we will find the mean and standard deviation of the observations kx₁, kx₂, kx₃, kx₄, kx₅, we get

So the mean of these 5 observations is

Substituting value from equation (i), we get

m₁=mk…………(iii)

And the standard deviation of the 5 observations is

Substituting equation (iii) in above equation we get

Substituting the value from equation (ii), we get

σ=ks

Hence the standard deviation of new set of observations, i.e., kx₁, kx₂, kx₃, kx₄, kx₅ is ks.

Given x₁, x₂, ... x_n be n observations

And Mean of these n observations,

And their standard deviation, SD_x=12.

Another series of n observations is given such that

w_i = lx_i + k for i = 1, 2, ...n, where l and k are constants

And mean of these n observations,

And their standard deviation, SD_w=15

Applying the given condition for mean we get

w_i = lx_i + k

Substituting the corresponding given values of means, we get

55 = l(48) + k…….(i)

Now we know

If standard deviation of x series is s, then standard deviation of kx series is ks,

So standard deviation of x₁, x₂, ... x_n is SD_x,

And hence the standard deviation of lx₁, lx₂, ... lx_n is lSD_x.

Similarly,

If standard deviation of x series is s, then standard deviation of k+x series is s,

So standard deviation of lx₁, lx₂, ... lx_n is lSD_x,

And hence the standard deviation of lx₁+k, lx₂+k, ... lx_n+k is lSD_x.

So applying the given condition for standard deviation, we get

SD_w=lSD_x

Substituting the given values, we get

15=l(12)

Now substituting the value of l in equation (i), we get

55 = (1.25)(48) + k

55=60+k

⇒ k=55-60=-5

Hence the values of k and l are -5 and 1.25 respectively

We know the standard deviation of the first n natural numbers is

Now for first 10 natural numbers, n=10, substituting this in the above equation of standard deviation we get

σ=2.87.

Hence the Standard deviations for first 10 natural numbers is 2.87

Now given numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

But these are first 10 natural numbers,

And we also know the standard deviation of the first n natural numbers is

Here n=10, substituting this in the above equation of standard deviation we get

Now when 1 is added to each numbers of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; we get new series as 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1.

Now we know, if standard deviation of x series is s, then standard deviation of k+x series is s,

So the standard deviation of 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1 series is also same as the standard deviation of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 series,

Hence

Now for variance we will square on both sides, we get

σ²=8.25

Hence the variance of the numbers so obtained is 8.25

First 10 positive integers are 1, 2, 3,4 ,5 ,6, 7, 8, 9, 10

on multiplying each number by – 1, we get

- 1, -2, -3, -4, -5, -6, -7, -8, -9, -10

on adding 1 to each of the number, we get

0, - 1, -2, -3, -4, -5, -6, -7, -8, -9

∴∑x_i = 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 = - 45

and

∑ x_i²= 0² + (- 1)² + (-2)² + (-3)² + (-4)² + … + (-9)²

But we know , so the above equation on applying this formula when n=9, we get

Now we know,

Substituting the corresponding values, we get

Now for variance we will square on both sides, we get

σ²=8.25

Hence the variance of the numbers so obtained is 8.25

Now we know standard deviation can be written as,

But given ∑x²=18000, ∑x = 960, N=60, substituting these corresponding values, we get

Now for variance we will square on both sides, we get

σ²=44

Hence the required variance is 44.

Given Coefficient of variation of two distributions are

CV₁=50 and CV₂=60

And there arithmetic means are

We know Coefficient of variation can be written as

Now for first distribution, we have

Substituting corresponding values, we get

⇒ σ₁=15………(i)

Now for second distribution, we have

Substituting corresponding values, we get

⇒ σ₂=15………(ii)

So from equation (i) and (ii), the difference of their standard deviation is 0.

given the standard deviation of some temperature data in °C, σ_C=5

We know that

Now we know

If standard deviation of x series is s, then standard deviation of kx series is ks,

So standard deviation of some temperature data in °C, σ_C=5,

And hence the standard deviation of some temperature data in ,

Similarly,

If standard deviation of x series is s, then standard deviation of k+x series is s,

So standard deviation of some temperature data in , ,

And hence the standard deviation of some temperature data in will be

Now for variance, we will square on both sides, we get

σ_F = 9²

=81

This is the required variance

Standard deviation

Explanation: We know Coefficient of variation can be written as

Where σ=standard deviation

=mean

zero, less than

Explanation: given is mean of n values, then the sum of all n terms is denoted by , so difference of both these is always equal to zero, i.e.,

And square of the above equation is also equal to zero, so

Now if f ‘a’ has the value other than , then

So,

So, if a has any value other than , then is less than.

11

Explanation: We know the square root of variance is standard deviation, i.e.,

σ² =121,

taking square root on both sides, we get

σ=√121=11

So, if the variance of a data is 121, then the standard deviation of the data is 11.

independent, dependent.

Explanation: Change of origin means some value has been added or subtracted in the observation.

And we know the standard deviation doesn’t change if any value is added or subtracted from the observations, so standard deviation is independent of change in origin.

But standard deviation is only affected by a change in scale, i.e., some value is multiplied or divided to observations.

Hence the standard deviation of any data is independent of any change in origin but is dependent of any change of scale.

minimum

Explanation: The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

least

Explanation: Mean deviation is sum of all deviations of a set of data about the data's mean.

And it is widely believed that the median is” usually” between the mean and the the mode.

So the mean deviation of the data is least when measured from the median

greater than or equal to

Explanation: we know standard deviation is difference between square of deviation of data about mean and square of mean.

And mean deviation is sum of all deviations of a set of data about the data's mean.

Hence, the standard deviation is greater than or equal to the mean deviation taken from the arithmetic mean