Uniform distribution Part-II and Normal distribution Part-I

# Uniform distribution Part-II and Normal distribution Part-I

hello and welcome to todays lecture so we
would like to continue and discuss about two very important distributions to today but
before i start i would like to briefly review what was discussed in last class ok so in
last lecture we discussed about two distributions the poisson distribution ok so for the poisson
distribution ok probability of x equal to i is given by e to the power minus lambda
lambda to the power i by factorial i i is equal to zero one two and it goes up to infinity
ok so poisson distribution the poisson random variable is a discrete random variable ok
but it can go up to very high values ok and what we had ah demonstrated in last class
was for the poisson distribution your expectation and variance both return you a value of lambda
ok and so poisson distribution also one more thing ah so for large n and small p so by
small i mean lets say p is order point one so we can have the binomial random variable
is almost equivalent to the poisson random variable with parameter lambda is equal to
n times ok so other than the poisson distribution the
other distribution we discussed yesterday was the uniform uniform distribution ok and
for a uniform distribution ok in a range alpha to beta your so your probability is uniform
ok its a constant and this constant so this c is nothing but c is given by ok one by beta
minus alpha ok so probability of a less than x less than b is nothing but b minus a by
beta minus alpha ok so let us discuss some few very simple cases so imagine you have
a random variable ok x which is uniformly distributed over zero to ten so lets say you
want to calculate probability of two less than x less than five what is this probability
so this is nothing as you know from this example your b is equal to five a is equal to two
so this probability is nothing but five minus two and beta and alpha what is beta is ten
alpha is zero so ten minus zero is equal to three tenth ok
similarly probability of lets say one less than x less than four is nothing will return
your same value ok two less than x less than five and one less than x less than four both
will return you a value of three by ten what is probability of x greater than six x greater
than six is nothing but since you can go from six to ten so x greater than six would be
ten minus six by ten minus zero is equal to four by ten ok so you see how you can make
use of the uniform distribution to calculate each of the probabilities ok
let us take one more example ok so imagine at the bus stop ok your buses arrive every
fifty minutes so they arrive at seven a m seven fifteen a m and so on and so forth ok
so seven thirty a m and so on and so forth ok so the question is so if a person arrives
at a time which is uniformly distributed between ok between seven a m and seven thirty
a m what is the probability of waiting less than five minutes ok so let us read the problem
so the buses arrive at every fifteen minutes so seven seven fifteen seven thirty so on
and so forth ok so the person arrives at a time ok which is a random variable and which
is uniformly distributed between seven and seven thirty which means that person has equal
chance of arriving at seven seven one so on and ah seven one ah up to seven thirty ok
so what is the probability of waiting less than fifteen minutes ok less than five minutes
ok so since your buses arrive at seven ok seven
fifteen then seven thirty so if he has to wait less than five minutes ok he can arrive
anytime between ten to fifteen ok or seven ten to seven fifteen or anytime between seven
twenty five to seven thirty ok so what is my probability so this probability ok waiting
probability of waiting less than five minutes is probability of ten less than x less than
fifteen plus the probability of twenty five less than x less than thirty ok this is what
this is five divided by thirty why because you have the entire duration is thirty minutes
and this is also five divided by thirty so total is ten by thirty equal to one third
ok now we can also ask the question what is the
probability of waiting more than twelve minutes ok so the bus arrives at seven seven fifteen
seven thirty ok so if the person has to wait more than twelve minutes then he has to arrive
greater than seven to seven three right because if he arrives at four then he has to wait
only for eleven minutes ok since he has to wait at least ok more ok actually waiting
at least so this if it is more than twelve minutes then you have seven to seven two ok
or if ah also seven three also ok and similarly he can any anytime between fifteen right after
fifteen to eighteen ok c if he arrives just after fifteen then he would
have to wait for thirty minutes which is more than twelve minutes and latest he can arrive
at seven eighteen in that case you will have to wait for exactly twelve minutes ok so my
probability is nothing but probability of zero less than x less than three plus probability
of fifteen less than x less than eighteen equal to three by thirty plus three by thirty
equal to six by thirty nothing but one fifth ok so for a uniform distribution ok once again
i right this is one by beta minus alpha for alpha less than x less than beta and zero
otherwise ok you can calculate the expectation expectation of this variable is nothing but
x this will return you a value of ok ok
expectation would returning a value of beta minus alpha by two similarly i can calculate
expectation of x square ok so x cubed by three and x cubed is this so you will get a value
of sorry sorry i think i have made a mistake ok so let me redo this ok so expectation of
x ok is x square so one by beta minus alpha ok x square by two between alpha to beta equal
to beta square minus alpha square by two into beta minus alpha ok is equal to alpha plus
beta by two ok that makes sense on an average you will get a value which is average of alpha
and beta ok this also i will redo so expectation of x square is going to be one minus beta
minus alpha into x cube by three ok is beta cubed minus alpha cubed ok beta square plus
alpha beta plus alpha square by three into beta minus alpha so this would give you a
value of beta square plus alpha beta plus alpha square into by three ok
so your variance of x then becomes alpha square by three minus alpha plus beta by two whole
square ok you can simplify this and see what value you get ok so that brings our discussion
of uniform distribution to a close we will now discuss one of the most important distributions
which is the normal distribution ok ok so what is the normal distribution so normal
distribution is the mount shaped distributions that you see at all times ok this is your
x value this is your probability ok so for normal distribution ok your f of x is given
by ok ok so you see there are two parameters
here so your probability density function is given by f x and one by sigma root two
pi to the power minus x minus mu whole square by two sigma square ok so typically this peak
is roughly point four by sigma square ok this peak is roughly sorry point four by sigma
ok so we want to know what is this value of mu and sigma respectively ok
so for that as again we can use the moment generating function so for this is equal to
expectation of e to the power t x this will be one i can take out this is a constant ok
ok so so phi of t then becomes ok i can let us say i put this mu minus y minus mu by sigma
is equal to sorry so let me define y as x minus mu by sigma ok so phi of t becomes sigma
root two pi ok so if i put this then my d y is equal to d x by sigma ok so i can actually
convert it ok so i have minus two e to the power t x e to the power minus x minus mu
whole square two sigma square into d x right so in this would mean my d x is equal to sigma
into d y so limits wont change but if i put this values again so d x by sigma is equal
to d y so this sigma goes the limits remain unchanged power sigma y plus mu ok y square
by two into d y ok so i can take into t ok so i can take this
mu t term out ok because the integral is over y ok and i can write it as minus infinity
to infinity e to the power ok minus y square minus two sigma ok plus y by two into d y
right so we have minus y square two minus ok y square by two minus two sigma ok so this
you can so minus two sigma is what you get ok ok so this is plus ok so this i can write
as e to the power ok mu t by root two pi ok i can write minus infinity to infinity exponential
so within the exponential i can write it as y minus sigma t whole square by two plus sigma
square t square pi two ok into d y ok so i can take out this term again so i can
write it as e to the power mu t plus sigma square t square by two ok by ok root two pi
can be separate so this is exponential together into one by root two pi minus infinity to
infinity e to the power minus y minus sigma t whole square by two d y ok so this term
this term is nothing but the same distribution f of x with parameter of mu is equal to sigma
t and sigma is equal to one ok so this integral gives you nothing but one ok so i can then
write the entire phi of t is nothing but exponential mu t plus sigma square t square by two ok
so phi prime t then becomes simply mu plus sigma square t into exponential of the same
thing ok so phi prime zero is equal to expectation
of x which will be giving me nothing but mu if you put t equal to zero here this will
return you a value of zero this will return you a value zero so exponential of zero is
one this will return you a value of zero so only you get this mu so expectation of x is
equal to mu phi double prime of zero ok so first let us find out phi double prime of
t is equal to ok phi double prime of t is equal to sigma square exponential of mu t
plus sigma square t square by two so basically i have taken a derivative of this term keeping
this as constant and plus into mu plus sigma square t exponential of mu t plus sigma square
t square by two ok so this will return you a so phi double prime
of zero will give me a value of sigma square ok plus so t equal to zero you only get a
value of mu here also you get a value of mu and this whole thing is one ok so you get
a value of sigma square plus mu square so this is nothing but e of x square ok so variance
of x so this we have already opted is nothing but e of x square minus e x whole square is
equal to sigma square plus mu square ok sigma square plus mu square minus mu whole square
nothing but sigma square ok so you get variance of x is nothing but sigma ok
so for a normal distribution the mean is equal to mu and the variance is equal to sigma square
ok and what you can also observe that from this distribution ok from this distribution
ok since this is symmetric about the center ok so this is also the mode y because this
is the maximum occurring value and this is also the median ok because this is right because
this is the symmetric distribution and for every point to the left there is another point
to the right which is occurring at identical frequency so their average will always give
you this particular value ok which is nothing so this is nothing but mu ok so it means that
for a normal distribution so mean is equal to median is equal to mode
ok now for this if i know that f of x is defined as one by sigma root two pi e to the power
minus x minus mu whole square by two sigma square right so i can define this variable
z as x minus mu by sigma then f x ok so this f x i can then find out a cumulative distribution
function so this becomes f of z ok becomes ok one by root two pi e to the power minus
z square by two ok and the cumulative distribution function
ok of phi of x ok phi of x then becomes one by root two pi into minus infinity to infinity
e to the power minus y square by two d y ok minus infinity to x ok minus infinity less
than x less than infinity ok so this is your value so what is y phi of
x for this distribution ok for any value x phi of x is this area under the curve ok this
is your total probability of finding any x in any any value less than x ok so we can
write down for the normal distribution ok so for probability of x less than b is nothing
but probability of z less than b minus mu by sigma ok is equal to phi of b minus mu
by sigma ok so this is the value so these values these values of phi are all been computed
and you can find them from tables existing tables ok in any statistics book ok so i can
write down so if this is probability of x less than b ok as i drew earlier so for for
any b this is my probability of x less than b so i can find out for the z variable what
is the correspondence so this correspondence to b minus mu by sigma and accordingly i can
look this up value up ok i can write probability of a less than x less than b is nothing but
phi of b minus mu by sigma minus phi of a minus mu by sigma ok
so one more thing because of symmetry ok if this is your distribution ok so this is for
the entire x for phi actually you will have mean is always equal to zero right so for
phi mean is always equal to zero ok so you have minus infinity to infinity so for any
value x right so this area will also be equal to this area ok in other words so phi of minus
x so lets say this is value of x and this is minus x so this is your z axis right so
phi of minus x you know this area and this area are exactly small ok so i can write phi
of minus x is equal to one minus phi of x ok so this whole area is unity entire integration
will return you a value of unity and this area is nothing but whole area minus this
area and this come comes because of symmetry ok so accordingly by looking up these phi
values i can find out any particular value where x has a value less than equal to one
ok less than equal to infinity ok so let us just take one sample example ok
so if x is a normal random variable with mu is three and sigma square equal to sixteen
we want to calculate probability of x less than eleven so how do i calculate i just define
z is equal to x minus mu by sigma in this case it is x minus three by four ok so p of
x less than eleven is nothing but probability of z less than eleven minus three by four
is equal to phi of two ok so you can find out this is by looking up this value what
is the value of probability of x less than by looking up the table ok
let us take another example ok lets say you have gasoline used in compact cars generally
ok have mileage mileage of thirty five miles per gallon with standard deviation of four
point five miles per gallon so if a company wants to make a car that out performance ninety
five percent ok of cars in the market ok what must be the mileage ok so essentially we have
been asked to ask for what value of x naught ok for what value of x naught do we get probability
equal to point nine five ok so what do we do we define z naught as x naught minus thirty
five by four point five ok and find out ok so we want to find out for what value of z
naught by looking up the table what value of z naught do we get probability of z naught
is point ninety five ok once we know once we know that so for so for
example if you look up the normal distribution tables you will see that z naught is roughly
one point six four five so then you can use this equation to invert and find out what
is the value of x naught ok so that ah completes our discussion of normal distribution so today
we briefly discussed about some cases of uniform distribution and then we came and described
about normal distribution so for our normal distribution we derived that your expectation
is equal to mu and your variance is equal to sigma square ok so you can define us another
normal distribution ok where you define so you rescale the axis so you define z equal
to x minus mu by sigma then you will get a distribution which is centered at zero so
it will have mean of zero and standard deviation of ok of minus ok with that you can so you
can use then you can use this particular distributions to scale and because you have tables which
have these values you can transfer any translate any variable x into a standard normal variable
z ok and then look up the table to find out what exactly is the value ok with that
i thank you for your attention and we will meet again in next class