The brain volumes (cm3) of 20 brains have a mean of 1106.4cm3 and a standard deviation of 123.2cm3. use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high. for such data, would a brain volume of 1402.8cm3 be significantly high?

Significantly high

Step-by-step explanation:

Given that the brain volumes (cm3) of 20 brains have a mean of 1106.4cm3 and a standard deviation of 123.2cm3

we are asked to use the given standard deviation and the range rule of thumb to identify the limits separating values that are significantly low or significantly high.

Since by rule of thumb we construct confidence interval 95% as 2 times std deviation on either side of the mean we find that

lower bound =

upper bound =

Hence any item less than 860 and any entry more than 1352.8 are considered as unusual.

or Significantly low or high.

Here 1402.8 cm^3 is higher than the upper bound hence significantly high

step-by-step explanation:

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answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.