00:01
This problem is talking about jim's systolic blood pressure, and we're assuming the given data has a normal distribution.
00:10
So therefore, everything we do will be based on the bell -shaped normal curve.
00:19
Jim's mean is 146, and his standard deviation is 22.
00:34
And in part a, if jim's systolic pressure is taken at random, what's the probability it will be 136 or less? so 136 would be to the right of the mean and we want to talk about being less.
00:54
So the probability that x is less than or equal to 136 is going to be comparable to x being less than 136.
01:07
So what we want to do is we want to find the z score associated with 136 by using the formula x minus mu over sigma.
01:17
So that would be 1306 minus 146 divided by 22, which is getting you a z score of negative 0 .45454545.
01:43
And it does say round your z score to two decimal places.
01:47
So we're going to utilize negative 0 .45.
01:51
And actually, i'm going to write it as negative 0 .45.
01:56
So if this z score is negative 0 .45, then the probability that x is less than 136 is comparable to the probability that z is less than negative 0 .45.
02:13
So at that point, you're going to utilize your standard normal table, which is appendix c2, and you'll find the units place and the 10th place along the side.
02:31
And you'll locate the hundreds place across the top.
02:37
And in doing so, you will find an area of 0 .32636.
02:44
Now, your table may only go out to four decimal places.
02:48
Your table might even only go out to three decimal places.
02:51
Regardless, you're going to get the same result.
02:54
So that represents the area to the left of that z score.
02:59
So this shaded area would be 0 .32636, which is what we are looking for.
03:10
So our probability that the blood pressure is less than or equal to 136, rounded to four decimal places, will be 0 .3264.
03:25
For part b, again, we have jim's systolic blood pressure, taken at random.
03:36
So again, we're going to have our bell -shaped curve.
03:40
His average was still 146, but this time we want the probability that it will be 176 or more...