00:01
In this question, we have given the waiting time for a license split renewal, is normally distributed with a mean of 30 minutes and standard deviation of 8 minutes, and an individual is randomly selected.
00:12
Let x be the waiting time for this person for the license split renewal.
00:18
So x follows the normal distribution, mean is 30, standard division is 8, so variance would be 8 square.
00:27
Part 1, want to find probability the individual will have a waiting time at least 10 minutes, that will be x greater equals to 10 and so if we were to just draw the normal distribution 30 is over here and 10 is here so we're looking at this portion here of probability so using the ti 84 calculator i'll be using normal cdf the lower limit would be 10 upper limit will be a large number 10 to power 99 the mean mule is 30, sigma is standard deviation and there'll be 8.
01:18
We click enter, you will get this, which is 0 .9938, 4 decimal place.
01:31
Part 2.
01:32
I want to find probability the person has a waiting time between 15 and 45 minutes.
01:38
So x is between 15 and 45.
01:50
We are interested in the red shaded part.
01:54
So again, using the normal cdf, the lower limit would be 15, upper limit will be 45.
02:07
Again, mew and sigma is the same at 30 and 8 respectively.
02:12
And so when click enter, you look at this.
02:19
There's 0 .932, 4 decimal place.
02:25
Part 3.
02:26
The director decides that 15 % of customers should receive a discount if they waited longer than predetermined time.
02:33
We want to find the number of minutes they need to wait to receive the discount...