00:01
Hello student here.
00:02
We have to answer the following question based on a probability distribution.
00:08
So here two assignment must be performed by same employee.
00:13
So here x is the minute complete to assignment one.
00:19
So here we have given mu x is equal to 20 and sigma x is equal to 5 now, why is the minute complete to assignment 2? so here we have given mu y which is equals to 30 and sigma y which is equals to 8 if x and y are normally distributed then and independent then what is the mean and standard deviation of time? so here mean is calculated as mu x plus mu y that is equals to 20 plus 30, which is equals to 50 to calculate standard deviation.
01:01
First we need to calculate the value of variance, which is calculated as 5 square plus 8 square that is 25 plus 64, which is equals to 89.
01:16
So here sigma is equal to square root of 89, which is equals to 9 .433.
01:25
Therefore the mean and standard deviation of time that is mu is equals to 50 and standard deviation sigma is equals to 9 .433.
01:37
Now in the next bit it is believed that probability of an employee on a particular absent is 0 .03 now assuming there are six people.
01:54
So here we have to find out what is the probability at least one is absent.
01:59
So here we use binomial distribution to calculate probabilities.
02:03
So here probability of x greater than 1 which is calculated as probability of 1 plus probability of 2 plus probability of 3 plus probability of 4 plus probability of 5 plus probability of 6.
02:21
So here 0 .15 plus 0 .01 plus 0 .00 plus 0 .00 plus 0 .00.
02:35
Now adding it will give us probability which is equal to 0 .16.
02:42
Now in the next bit here we have given event a and b are independent and we have given probability of a which is equal to 0 .2 and we have given probability of b, which is equal to 0 .6...