00:02
Right, so according to the problem, we're looking at female's height here.
00:14
Okay, so the me is equal to 63 .7.
00:25
So, 63 .7 and we have standard deviation that is equal to 2 .9.
00:34
Now we're as if we have a raw score, the percentage of the female pilots having a high between 62 and 78.
00:54
So if we're looking at the standard normal curve, assuming that the data is normal, we're actually looking for the area under this curve.
01:06
That's 62 to 78.
01:10
So in order to do that, the most standardize our x values from x to z.
01:17
Okay, so we use the formula z equals x minus mu all over sigma.
01:26
So if the x is 62, we have z equals 62 minus 63 .7 over 2 .9.
01:38
Okay, and then if d x is equal to 7 to 8, we have 7 to 8 minus 63 .7 over 2 .9.
01:48
So we just have to calculate these values.
01:53
Okay, so allow me to calculate.
02:05
So calculating that we have a z score for z score that's right here, negative 0 .5816.
02:19
And then another z, okay.
02:31
So another z that's equal to 4 .9310.
02:40
Okay.
02:41
So in a standard normal curve, we're actually looking for the probability that the z is between negative 0 .5862 to 4 .9310.
02:56
Okay, so in order to do that, we have to take note that this probability is equal to the probability that z is less than 4 .0.
03:12
Sorry.
03:16
Z is less than 4 .930 minus the probability that z is less than negative 0 .5862.
03:29
Okay, so just have to look at the z table in order to get its values.
03:43
Okay, let me scroll this one out.
03:46
Okay, so the probability that z is less than 4 .930 is, okay, it's around one.
04:07
Okay, it's almost covering all the area.
04:11
Okay, minus the probability that, okay, z less than negative 0 .5862.
04:26
This is 0 .27, okay, 27.
04:33
8 .89.
04:36
Okay.
04:38
So calculating this one and converting 2%, okay.
04:43
So we have 1 minus 2 .7 weekline.
04:51
We have 0 .7 to 11.
04:57
Multiply it by 100%.
04:59
We have 72 .11%...