00:01
So this one's a long one.
00:03
Next time you might want to break it up a little bit.
00:06
But in question three, it says we're normally distributed that i'm going to call nd with a mean of 4 .7, and we don't know the standard deviation.
00:17
But we're told that the probability that x is less than 4 .5 is 0 .3075.
00:25
And the probability that x is greater than 5 .2 is 0 .1046.
00:34
And we know z is the data item minus the mean divided by the standard deviation.
00:40
Well, if we know this, it doesn't matter which one we choose, i'm using this one.
00:50
Well, for z to be 10 .46, z is 1 .146.
00:55
Z is 1 .5.
01:07
2557, which means we can use that to solve for standard deviation with the z score.
01:14
So we could solve 1 .2557 equal to 5 .2 minus the mean is 0 .5, divided by the standard deviation.
01:25
We get the standard deviation to be 0 .3998, 398, 398, if i can spit it.
01:37
Out.
01:39
So in part a, when we want the probability that 4 .5 is less than x is less than 5 .2, this one i don't need the standard deviation.
01:48
I could say one minus the probability that x is less than 4 .5, minus the probability that x is greater than 5 .2 and get 0 .587.
01:59
For b knowing the standard deviation is helpful because i want the probability that x is between 4 .2 and 4 .9 which is the probability that z is between negative 1 .257 and 0 .5028 which is 0 .587 c, the probability that x is greater than 4 .9, is the probability that z is greater than 0 .50228, which is 0 .3077353.
02:53
And d, the probability that x is less than 4 .2 is the probability that z is less than negative 1 .277...