Easiest method you can apply (Cramer's one). <em>"</em><em>Thanks</em><em> </em><em>me</em><em>"</em><em> </em>if i've been helpful
if ashley is in the 85th percentile and the mean of her test scores is 80 with a standard deviation of 15, which of the followin
g scores would ashley most likely get on a standardized test?
2 answers:
The 85th percentile is a probability of 0.85 and its location is shown in the bell curve below
We need to look up on the z-table the value of z when P(Z<z) = 0.85 or P(Z>z) = 0.15
Value of z-score when p-value P(Z>z) = 0.15 is z = 1.0364
Next step is to use the formula of z-score to work out X
z-score = (X-μ) / σ
1.0364 = (X-80) / 15
1.0364 × 15 = X - 80
X = 15.546 + 80
X = 95.5
Most likely score that Ashley gets is 95.5
We need to look up on the z-table the value of z when P(Z<z) = 0.85 or P(Z>z) = 0.15
Value of z-score when p-value P(Z>z) = 0.15 is z = 1.0364
Next step is to use the formula of z-score to work out X
z-score = (X-μ) / σ
1.0364 = (X-80) / 15
1.0364 × 15 = X - 80
X = 15.546 + 80
X = 95.5
Most likely score that Ashley gets is 95.5
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Well, what I would do [not sure if it's the correct way], is use the white area as a sector of a circle. The top right of the box is the center of the circle, so the radius is 6.
Then, by the white portion being one quarter of the circle (90° out of 360°), I could calculate the that pink shaded region = the total square (6×6) minus the white sector.
So area (A) of white sector (s): A(s) = 1/4×pi×r^2
36 - 28.27 = 7.73
Then, by the white portion being one quarter of the circle (90° out of 360°), I could calculate the that pink shaded region = the total square (6×6) minus the white sector.
So area (A) of white sector (s): A(s) = 1/4×pi×r^2
36 - 28.27 = 7.73