GRE Exam > GRE Videos > Mathematics for GRE Paper II > Calculating Standard Deviation for Discrete and Continuous Frequency Distributions
Calculating Standard Deviation for Discrete and Continuous Frequency Distributions Video Lecture | Mathematics for GRE Paper II
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FAQs on Calculating Standard Deviation for Discrete and Continuous Frequency Distributions Video Lecture - Mathematics for GRE Paper II
| 1. How is the standard deviation calculated for a discrete frequency distribution? |  |
Ans. To calculate the standard deviation for a discrete frequency distribution, you need to follow these steps:
1. Calculate the mean of the distribution by multiplying each data value by its respective frequency, summing them up, and then dividing by the total frequency.
2. Subtract the mean from each data value and square the result.
3. Multiply each squared difference by its respective frequency and sum them up.
4. Divide the sum of the squared differences by the total frequency.
5. Take the square root of the result obtained in step 4 to get the standard deviation.
| 2. How is the standard deviation calculated for a continuous frequency distribution? | |
Ans. To calculate the standard deviation for a continuous frequency distribution, you need to follow these steps:
1. Calculate the mean of the distribution by multiplying each data value by its respective frequency, summing them up, and then dividing by the total frequency.
2. Subtract the mean from each data value and square the result.
3. Multiply each squared difference by its respective frequency and sum them up.
4. Divide the sum of the squared differences by the total frequency.
5. Take the square root of the result obtained in step 4 to get the standard deviation.
| 3. What is the importance of calculating the standard deviation in frequency distributions? | |
Ans. The standard deviation is an essential measure in frequency distributions as it provides information about the dispersion or spread of the data. A higher standard deviation indicates that the data values are more spread out from the mean, while a lower standard deviation suggests that the data values are closely clustered around the mean. By calculating the standard deviation, we can quantify the variability within the data and make comparisons between different distributions.
| 4. Can the standard deviation be negative? | |
Ans. No, the standard deviation cannot be negative. It is always a non-negative value or zero. The standard deviation represents the average amount of deviation or dispersion from the mean. As deviations are squared during the calculation process, the result is always positive. Thus, the standard deviation provides a measure of spread that is always non-negative.
| 5. How does the standard deviation relate to the shape of a frequency distribution? | |
Ans. The standard deviation is closely related to the shape of a frequency distribution. It helps determine whether the distribution is symmetric or skewed. In a symmetric distribution, the mean, median, and mode are approximately equal, and the standard deviation provides a measure of how close the data values are to this central tendency. In a skewed distribution, where the mean, median, and mode differ, the standard deviation can help identify the extent of the skewness. A higher standard deviation in a skewed distribution indicates a wider spread on one side of the distribution.
Text Transcript from Video
we have to compute standard deviation
from the following distribution we have
class and the frequency so what we
understand is this is the case of
continuous frequency distribution okay
so in case of continuous frequency
distribution we have two approaches for
finding standard deviation two formulas
are available let us make use of the
first formula and then we'll see how
this goes the first formula that is
available is standard deviation Sigma is
square root of summation if I D I square
okay
divided by n minus summation F I D I by
n the whole square so how we are going
to find D how we are going to find n
that's what we are going to discuss
so first let's focus on D so how will
find D what is the approach D is nothing
but X I minus a so what is X I X I is
nothing but the midpoint of your class
intervals and ya is any convenient value
so first let's find X I X I is going to
be the midpoint of this class interval
that is ten plus nineteen by two
fourteen point five so this is the
midpoint for the first class interval 20
plus 29 by two we get 24 point five then
30 plus 39 by two we get thirty four
point five then 40 plus 49 by two it is
forty four point five and 50 plus 50 9
by two it's going to be fifty four point
five so this is X and we can find D AI
di is X minus a a can be any convenient
value so I'm going to pick up this 34.5
as a so di is X I minus a so it is
fourteen point five minus 34 point five
I get minus twenty twenty four point
five minus thirty four point five I get
minus 10 then the third deviation that
is thirty four point five - yay that got
to be zero then forty four point five
minus thirty four point five we get ten
and fifty four point five minus thirty
four point five we get twenty so this is
deep but to looking at the formula we
need d square then FD square and F into
D okay so let's have d square let's have
d square that is minus twenty square
that's going to be four hundred minus 10
squared is 100 0 100 and 400 so this is
d square then we need F D so we'll have
F I D I so here we have the frequency
which is F I F I & bi 8 into 400 is that
right let's make a cross verification
now what is the F here we are right in
picking up F but this is d square what
is relevant for us is D I right so it is
8 into 20 so that's going to be 160 but
it's a negative number minus 160 then 14
into 10 minus 140 then 20 into 0 it's
going to be 0 22 into 10 it's going to
be 220 and lovin in 220 it's going to be
again 220 so this takes care of FB but
we also need F I di square we also need
F I di square so this di square got to
be multiplied with this frequency so 8
into 400 it's going to be 3,200 they
14 in 200 that's thousand four hundred
then the next item any hope it should be
0 22 in 200 is 2200 and final lovin into
400 that should be four thousand four
hundred so we have all values in place
now let's try to fit them in formula the
first item let me just pick up the first
item is summation F I D I square it
means I have to total this so it's going
to be summation F I di square and if I
total this I should be getting loud
thousand two hundred then I need excuse
me then I need summation if I di okay so
this is summation F I D I and if I do
the totalling here it's going to be one
hundred and forty then I need
information about N in a simple in is
summation of F ok n is summation of F
and this is F if I do that I get
summation of F which is N and that's
going to be 75 so let me capture that as
well
now I can fit all this in the formula
let me do that standard deviation is
square root of summation fi di square
that is 11 200 divided by n which is 75
minus summation FD square that is 140
divided by 75 the whole square so if I
do this calculation eventually I'll get
standard deviation is equal to twelve
point zero eight so this is one example
for us to understand the standard
deviation in case of continuous
frequency distribution let's make use of
the other formula and we'll see how to
find standard deviation in
next session
from the following distribution we have
class and the frequency so what we
understand is this is the case of
continuous frequency distribution okay
so in case of continuous frequency
distribution we have two approaches for
finding standard deviation two formulas
are available let us make use of the
first formula and then we'll see how
this goes the first formula that is
available is standard deviation Sigma is
square root of summation if I D I square
okay
divided by n minus summation F I D I by
n the whole square so how we are going
to find D how we are going to find n
that's what we are going to discuss
so first let's focus on D so how will
find D what is the approach D is nothing
but X I minus a so what is X I X I is
nothing but the midpoint of your class
intervals and ya is any convenient value
so first let's find X I X I is going to
be the midpoint of this class interval
that is ten plus nineteen by two
fourteen point five so this is the
midpoint for the first class interval 20
plus 29 by two we get 24 point five then
30 plus 39 by two we get thirty four
point five then 40 plus 49 by two it is
forty four point five and 50 plus 50 9
by two it's going to be fifty four point
five so this is X and we can find D AI
di is X minus a a can be any convenient
value so I'm going to pick up this 34.5
as a so di is X I minus a so it is
fourteen point five minus 34 point five
I get minus twenty twenty four point
five minus thirty four point five I get
minus 10 then the third deviation that
is thirty four point five - yay that got
to be zero then forty four point five
minus thirty four point five we get ten
and fifty four point five minus thirty
four point five we get twenty so this is
deep but to looking at the formula we
need d square then FD square and F into
D okay so let's have d square let's have
d square that is minus twenty square
that's going to be four hundred minus 10
squared is 100 0 100 and 400 so this is
d square then we need F D so we'll have
F I D I so here we have the frequency
which is F I F I & bi 8 into 400 is that
right let's make a cross verification
now what is the F here we are right in
picking up F but this is d square what
is relevant for us is D I right so it is
8 into 20 so that's going to be 160 but
it's a negative number minus 160 then 14
into 10 minus 140 then 20 into 0 it's
going to be 0 22 into 10 it's going to
be 220 and lovin in 220 it's going to be
again 220 so this takes care of FB but
we also need F I di square we also need
F I di square so this di square got to
be multiplied with this frequency so 8
into 400 it's going to be 3,200 they
14 in 200 that's thousand four hundred
then the next item any hope it should be
0 22 in 200 is 2200 and final lovin into
400 that should be four thousand four
hundred so we have all values in place
now let's try to fit them in formula the
first item let me just pick up the first
item is summation F I D I square it
means I have to total this so it's going
to be summation F I di square and if I
total this I should be getting loud
thousand two hundred then I need excuse
me then I need summation if I di okay so
this is summation F I D I and if I do
the totalling here it's going to be one
hundred and forty then I need
information about N in a simple in is
summation of F ok n is summation of F
and this is F if I do that I get
summation of F which is N and that's
going to be 75 so let me capture that as
well
now I can fit all this in the formula
let me do that standard deviation is
square root of summation fi di square
that is 11 200 divided by n which is 75
minus summation FD square that is 140
divided by 75 the whole square so if I
do this calculation eventually I'll get
standard deviation is equal to twelve
point zero eight so this is one example
for us to understand the standard
deviation in case of continuous
frequency distribution let's make use of
the other formula and we'll see how to
find standard deviation in
next session
About this Video
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Dec 25, 2024 Last updated |
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