What is the slope for 3y+6x=9
1 answer:
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The variable is
X: blood platelet count of a woman
This variable has a bell shaped distribution (Normal)
With mean μ=261.1 and standard deviation σ=64.3
a.
The platelet count within 3 standard deviations of the mean can be calculated as
μ±3σ
You can symbolize it as
μ-3σ ≤ X ≤ μ+3σ
According to the empirical rule of the normal distribution you know that under any bell shaped distribution:
Between μ ± σ you'll find the 68% of the distribution
Between μ ± 2σ you'll find the 95% of the distribution
Between μ ± 3σ you'll find the 99% of the distribution
So following this rule you'll find 99% of the women with platelet count within 3 standard deviations of the mean.
b.
3% of the population is found between a vertain interval, the rest 97% is separated in equal tails around this interval.
Divide 0.97 by 2
0.97/2= 0.485
Each tail contains 0.485 of the distribtuion.
a and b represent the values within you'll find 3% of the population.
Until a you'll find 0.485 of the population and until b you'll find 0.485+0.03=0.515
To calculate both unknown values you have to use the standard normal distribution. This distribution is centered in zero, the left tail is negative and the right tail is positive.
a and b are at a equal distance from the mean but with different sign, so you only have to calculate on and then you invert the sign to get the other one.
I'll calculate the positive value:
Z= (X- μ )/σ~N(0,1)
P(Z≤ b)=0.515
Look in the body of the Z-table, rigth or positive entry and reach the margins for the value:
b= 0.038
Then using the values of the mean and the standard deviation you can calculate the corresponding valueof platelet count:
b= (X- μ )/σ
0.038= (X- 261.1 )/64.3
0.038*64.3=X-261.1
2.4434+261.1=X
X=263.5434=263.5
a=-b=-0.038
a= (X- μ )/σ
-0.038= (X- 261.1 )/64.3
-0.038*64.3=X-261.1
-2.4434+261.1=X
X=258.6566= 258.7
3% of the populations platelet count is between 258.7 and 263.5.
Answers:
- Part a)
- Part b) 3
- Part c) 2.24
- Part d) 1.58
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Work Shown:
Part (a)
The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)
The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3
We apply the distance formula to say...
We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.
Saying is the same as writing d = sqrt(x^2-(x^2-3)^2)
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Part (b)
Plug in x = 0 and you should find the following
This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).
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Part (c)
Repeat for x = 1
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Part (d)
Graph the d(x) function found back in part (a)
Use the minimum function on your graphing calculator to find the lowest point such that x > 0.
See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.
The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)
This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.
