Question 1143106: I am using Excel as my choice of technology. I am requesting assistance with how to calculate these problems:
According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is normally distributed.
d. Find the probability that a person in China has blood pressure between 120 and 125 mmHg. Used Excel for computation =SUM(0.36399)+(0.55189)-1=-0.084126- the online calculator is giving me positive number but excel is giving me a negative number.
e.What blood pressure do 90% of all people in China have less than?-How do I even begin to calculate this?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! from what i can see, .....
you got the area to the left of a z-score of -.34783 = .363985.
you got the area to the right of a z-score of -.13043 = .551889.
you then derived the area to the left of a z-score of -.13043 by taking .551889 - 1 and getting -.448111.
you did that wrong, because you should have gotten the area to the left of a z-score of -.13043 = 1 - .551889 = .448111.
your formula was SUM(0.363985)+(0.551889)-1
if i calculate that out, i get .363985 +.551889 - 1 = -.084126
when you get the area of the interval, you take the larger area and subtract the smaller area from it.
in both cases, you get the area to the left of the z-score.
area to the left of a z-score of -.34783 is .363985
area to the left of a z-score of -.13043 = .448111.
you then subtract .363985 from .448111 which becomes .084126.
sincne the normsdist function give you the area to the left of the z-score, you should have just gotten normsdist(-.34783) = .363985 and normsdist(-.13043) = .448111.
you then subtract the smaller area from the larger are to get the area in between.
in your second problem, you want to know the blood pressure that 90% of the people in ch in china have less than.
you find the z-score that has an area of .9 to the left of it.
in excel, that would be the using the normsinv formula as shown in the attached excel spreadsheet.
you would get normsinv(.9 = 1.281552.
it is telling you that the z-score that has 90% of the area under the normal distribution curve to the left of it is 1.281552.
you then use that z-score to get the raw score.
the z-score formula is z = (x - m) / s
z = 1.281552
x = x
m = 128
s = 23
you get 1.281552 = (x - 128) / 23
multiply both sides of this equation by 23 and add 128 to both sides of this equation to get:
1.281552 * 23 + 128 = x
solve for x to get x = 157.475696
what you see in excel will probably be a little different because of differences in rounding, but it should be close to that number.
there is an online calculator that i use that give you graphical displays of that's going on.
that calculator can be found at http://davidmlane.com/hyperstat/z_table.html
with a mean of 128 and a standard deviation of 23, this calculator can tell you the area between 120 and 125 directly as shown below.
with a mean of 128 and a standard deviation of 23, this calculator can tell you the raw score that has an area of .9 to the left of it as shown below.
i find excel not as friendly as other tools.
most of the time i use my TI-84 Plus.
regardless of what tool you use, you have to keep track of the formulas and the concepts in order to get the right answer.
area in between 2 z-score is the area to the left of the larger z-score (more positive) minus the area to the left of the smaller z-score (less positive).
|
|
|
