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4 years ago

What is the domain and range for the following function an its inverse? f(x)=-x+5

1 answer:

5 0

The function and its inverse are exactly the same. You find the inverse of a function by switching the x and the y and then solving for the new y. If your function is y = -x + 5 and you switch the x and the y, you get x = -y +5. Now solve for the new y to get y = -x + 5. The domain and range for those is all real numbers.

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According to the 2017 Corporate Travel Index compiled by Business Travel News, the average daily cost for business travel in the

Gemiola [76]

Answer:

1) X[bar]= $262.60

2) S= $83.55

3) [$57.47; $152.55]

Step-by-step explanation:

Hello!

The objective is to study the average daily cost for business travel in the United States. To test this a sample of the estimated living expenses to 10 international cities was taken. The study variable is X: estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.

Sample data:

n=10

242.87, 260.93, 194.19, 260.86, 355.36, 211.00, 284.08, 139.16, 436.72, 240.87

∑Xi= 2626.04

∑Xi²= 752439.39

The sample mean is calculated by adding all the observed costs and divide it by the sample size:

X[bar]= ∑Xi/n = 2626.04/10= $262.60

To calculate the standard deviation of the sample you have to calculate the sample variance first using the formula:

S²= $\frac{1}{n-1}[∑Xi^2-\frac{(∑Xi)^2}{n} ]$

S²= $\frac{1}{9}[752439.39-\frac{(2626.04)^2}{10} ]$

S²= 6981.1979

S= √S² = √6981.1979 = $83.55

To calculate a Confidence Interval for the population standard deviation, you have to first calculate the Confidence Interval for the population variance and then calculate it's square root to obtain the interval for the standard deviation.

This happens because to study a population parameter you need a statistic that follows the following conditions: 1) have known distribution, 2) include the population parameter and it's point estimator, 3) the only incognito of the statistic equation must be the parameter under study. There is no statistic that meets these conditions, but the chi-square statistic meets these conditions and allows you to study the population variance. And this is only possible if the study variable has a normal distribution.

Assuming that the daily living costs have a normal distribution, the formula of the Confidence Interval for the population variance is:

$[\frac{(n-1)S^2}{X^2_{(n-1); 1- \alpha/2} } ; \frac{(n-1)S^2}{X^2_{(n-1); \alpha/2} } ]$

$[\frac{9*6981.1979}{19.023} } ; \frac{9*6981.1979}{2.700} } ]$

[$²3302.88; $²23270.66]

So with a confidence level of 95% you'd expect that the confidence interval

[$²3302.88; $²23270.66] will contain the population variance of the estimated living costs (including single room at 4 stars hotel, beverages, breakfast, taxi fares and incidental costs) on an international city.

Now you calculate the square root of the interval:

[√$²3302.88; √$²23270.66]

[$57.47; $152.55]

And at the same level, you'd expect that the interval [$57.47; $152.55] contains the population standard deviation of the estimated living costs (including a single room at 4 stars hotel, beverages, breakfast, taxi fares, and incidental costs) on an international city.

I hope it helps!

6 0

3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$