Help? Statistics...

Mathematics

2 answers:

BartSMP [9]3 years ago

7 0

On your calculator, use the normalcdf() function, enter the Lower Limit, Upper Limit, Mean, and Standard deviation.

The mean and standard deviation are given to you. You want to find "less than 13.5 seconds", meaning that 13.5 is the maximum "Upper Limit" and 0 would be the Lower Limit because we can pretend 0 second is the minimum. Don't worry about negatives, in this scenario, values like "-5 seconds" are impossible.

Enter normalcdf(0, 13.5, 13.56, 2.24) in a calculator and the output result is the answer.

Irina-Kira [14]3 years ago

4 0

Mean = μ =13.56; Standard deviation = σ = 2.24.
Let's consider the Z distribution where μ =0 and σ = 1
(with x = 13.50)
Let's calculate the Z score: Z = (x-μ)/σ .

Z = (13.50 - 13.56)/2.24 → Z = - 0.026785

For Z = -0.02 , P(Z) = 0.4920 OR 49.20%
For Z = -0.03 , P(Z) = 0.4880 OR 48.80%
Then Z (Approx) = [(48.2)+(48.8)]/2 = 49% (approx.)

Then 49% of a runner chosen at random will have a 200-meter time less than 13.5 seconds

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Write an integral in polar coordinates equivalent to the following:

Alex73 [517]

The region is a segment of a circle with radius √2 lying above the line $y=1$ . In polar coordinates, this line has equation

$y = r\sin(\theta) = 1 \implies r = \csc(\theta)$

and the circle has equation

$x^2+y^2 = r^2 = 2 \implies r=\sqrt2$

The two curves meet when

$\csc(\theta) = \sqrt2 \implies \sin(\theta) = \dfrac1{\sqrt2} \implies \theta = \dfrac\pi4 \text { or } \theta = \dfrac{3\pi}4$

Then the same integral in polar coordinates is

$\displaystyle \int_{-1}^1 \int_1^{\sqrt{2-x^2}} xy \, dy \, dx = \boxed{\int_{\pi/4}^{3\pi/4} \int_{\csc(\theta)}^{\sqrt2} r^3 \cos(\theta) \sin(\theta) \, dr \, d\theta}$