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12 days ago

7

The manager of a healthy food store has determined that the weekly demand for a popular type of granola is a normally distribute

d random variable with mean 85 pounds and standard deviation 5 pounds. If the demand for a given week falls within the lowest 2.5% of all possible values for the weekly demand, the price of the granola will be reduced for the following week. Calculate the value in pounds (lbs) for the weekly demand below which the manager will have to reduce the price.

1 answer:

babunello [11.8K]12 days ago

8 0

The weight is 75.2 pounds. The analysis utilizes the z-score formula addressing normally distributed samples to discern that the weekly granola demand, when below the lowest 2.5%, leads to a price reduction.

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Step-by-step explanation:

The equation representing the sphere, which has its center at the origin, can be written as $x^2+y^2+z^2 = 64$ . For z equal to 4, we find

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c) We will build on the analysis from earlier to set limits in both Cartesian and polar coordinates. Initially, we recognize that x spans from $-4\sqrt[]{3}$ to $4\sqrt[]{3}$ . This determination is made by fixing y = 0 and identifying the extreme x values that fall on the circle. For y, we observe that it ranges between $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , which holds because y must reside within the interior of the identified circle. Lastly, z will extend from 4 up to the sphere; hence, it varies from 4 to $\sqrt[]{64-x^2-y^2}$ .

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$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Remember that the cylindrical coordinates are expressed as $x=r\cos \theta, y = r\sin \theta,z = z$ , where r denotes the radial distance from the origin projected onto the x-y plane. Also note that $x^2+y^2 = r^2$ . We will derive new limits for each of the transformed coordinates. Recall that due to the prior circular constraint, $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ varies between 0 and $2\pi$ . Furthermore, r starts from the origin and extends to the edge of the circle, with r reaching a maximum of 4\sqrt[]{3}. Lastly, Z increases from the plane z=4 up to the sphere, where it is constrained by \sqrt[]{64-r^2}. Thus, the integral that computes the desired volume is as follows:

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . It’s important to note that the r factor arises from the Jacobian associated with the transition from Cartesian to polar coordinates, ensuring the integral maintains its value. (Explaining how to calculate the Jacobian exceeds the scope of this response).

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