00:01
Now, let's solve this problem.
00:02
In this problem, mu value that is mean is given as mu and that will be n multiplied p.
00:08
So since this is a problem of binomial distribution, the n value will be my sample size which is given as 400 and p value is the probability of success which is given as 0 .30.
00:19
That is 30%.
00:20
So new comes out to be 120.
00:23
Right.
00:24
Now if i write the value of sigma, that is standard deviation, then it will be n multiplied p multiplied q so what is a value of q you is nothing but one minus p that is q is the you know probability of failure so okay so q is one minus p so i hope this value is known to you then if i just plug in the value like 400 multiplied 0 .3 0 multiplied 0 .70 then what will happen you will have your value for standard deviation as 9 .1 so, therefore, if i just go on and write that probability of x greater than equal to one one zero will be equals to one minus probability of x greater than or it will be probability of x less than.
01:17
So x less than or equal to 110.
01:21
Okay, so here what i have used is that, i have used this formula that if i have probability of x greater than a, then it is a.
01:29
Will be equals to 1 minus probability of x less than a okay so using this very concept i can write the value as for the above part okay so i'm just i'm particularly solving this this particular equation now the value will come around 1 minus probability of how much will be the value inside the bracket it will be x minus 120 over how much will it be at the denominator 1 9 .17 and then it will be less than equal to 1110 minus 120 over 9 .17.
02:06
Okay, right.
02:07
So, if you will solve it further, you can, you know, write this one as upon solving that this will be equals to and that z will be 1 .09 and then here it will be negative of 1 .0...