00:01
Hi, from the first screen given that here x represents the score in examination and it is normally distributed with mean 80 and the standard deviation is 4.
00:18
So in part a we need to find the probability of 68 less than x is less than 92.
00:26
Then z is equal to x minus mu divided by sigma.
00:31
So probability of 68 minus mu divided by sigma is less than x minus mu divided by sigma is less than 92 minus mu divided by sigma.
00:45
So that is equal to probability of 68 minus 80 divided by 4 is less than z is less than 92 minus 80 divided by 4.
00:56
So that is equal to probability of minus p is less than z is less than 3.
01:03
By using the standard normal distribution table we obtain 0 .9973.
01:09
Hence we conclude that the percent of individual scoring between 68 and 92 will be 99 .73%.
01:31
Now let us move on to part b.
01:32
So in part b we need to find probability of scoring between 72 is less than x is less than 84.
01:44
So as representing the same method so we obtain 72 minus 80 divided by 4 is less than z is less than 84 minus 80 divided by 4.
01:59
So that is equal to probability of minus 2 is less than z is less than 1.
02:07
By using the standard normal distribution table we obtain 0 .8186.
02:15
Hence we conclude that the percent of individual scoring between 72 and 84 will be 81 .86%.
02:45
Now let us move on to part c.
02:46
So in part c we need to find the probability of x is greater than 84.
02:53
So that is equal to 1 minus probability of x is less than or equal to 84...