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Mashutka [201]
3 years ago
13

A retired auto mechanic hopes to open a rustproofing shop. Customers would be local new-car dealers. Two locations are being con

sidered, one in the center of the city and one on the outskirts. The central city location would involve fixed monthly costs of $6,980 and labor, materials, and transportation costs of $30 per car. The outside location would have fixed monthly costs of $4,200 and labor, materials, and transportation costs of $40 per car. Dealer price at either location will be $90 per car.
a.

Which location will yield the greatest profit if monthly demand is (1) 200 cars? (2) 300 cars?
200 cars: (City or Outside?) yields the greatest profit.
300 cars: (City or Outside?) yields the greatest profit.

b.

At what volume of output will the two sites yield the same monthly profit?

Volume of output = _______cars

Mathematics
1 answer:
const2013 [10]3 years ago
3 0

It is convenient to let a spreadsheet or graphing calculator do the math for this. Functions can be defined for cost and profit, and evaluated at each of the volumes of interest.

a1) For 200 cars, the Outside location yields the greatest profit

a2) For 300 cars, the City location yields the greatest profit

b) The sites yield the same profit for a volume of 278 cars.

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Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the
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Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

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So, the sample mean is 31.3.

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\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}

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\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\

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\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98

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A study investigating the relationship between age and annual medical expenses randomly samples 400 individuals in a city. It is
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Answer:

The probability is 0.789

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\sigma = 16, n = 400, s = \frac{16}{\sqrt{400}} = 0.8

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