00:01
In this question, we have been asked to calculate the value of probability such that mu is here equals to we can say 14.
00:07
Now here the value of sigma will be here 3 .8 and now the value of n that is sample size will be here 65.
00:15
So the value of probability such that x bar is here minus 14 and then it will be here lesser than 1.
00:23
So this is going to be here we can say probability of minus 1 and then it will be here less than x bar minus 14 and then here it is 1.
00:32
So this very value can be here written as we can say it will be probability of minus 1 minus 14 and then here it is going to be x bar and then it will be here we can say lesser than 1 plus 14.
00:47
So now using the information that the z is here equals to we can say x bar minus mu x and then here it is we can say sigma x bar and now it will be here equals to we can write it as x bar minus mu and then in the denominator we are going to have sigma divided by under root of n.
01:09
So using this very information we can here say that z1 value will be here equals to minus 13 and then here it is minus 14 now it will be here 3 .8 and then here it is under root of 65.
01:23
So this is here going to be equals to negative 2 .12 and this is the approximate value and hence we can say that the z2 value will be here 15 minus 14 and then in the denominator we are going to have 3 .8 divided by 65 and this will be here equals to we can say it is also going to be here 2 .12 and hence the above expression can be here simplified and it will be here written as probability of negative 2 .12 lesser than z and that is lesser than 2 .12.
01:59
So this will be here equals to we can say area between negative 2 .12 and positive 2 .12.
02:06
So this can be here written as probability of z and that is lesser than 2 .12 and then it will be here we can say minus probability of z and that is here lesser than negative 2 .12.
02:19
So this very information can be here used to write the value as probability of negative 2 .12 and then here it is lesser than z and that is here lesser than 2 .12.
02:31
So this will be here equals to 0 .9830 and then it is here minus 0 .0170.
02:39
So this very value will be here equals to we can write it as 0 .9660.
02:46
So we have obtained the required value of probability and it can be here put inside a box in order to highlight it and it will here look like this.
02:56
So now let us see how we can obtain the value of probability in the second part of this very problem.
03:01
So in the second part we have to calculate the probability value such that it is here going to be we can say x bar minus 14 and then it will be here lesser than 0 .8 and then here it is 333 and then it will be here 33.
03:17
So here the value of mu is equals to 14, the value of sigma is equals to 3 .8 and the value of n will be here equals to 65...