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solniwko [45]
3 years ago
14

Suppose monthly rental prices for a one-bedroom apartment in a large city has a distribution that is skewed to the right with a

population mean of $880 and a standard deviation of $50.
(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?
(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?
Mathematics
1 answer:
omeli [17]3 years ago
8 0

Answer:

a) Nothing, beause the distribution of the monthly rental prices are not normal.

b) 1.43% probability that the sample mean rent price will be greater than $900

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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

(a) Suppose a one-bedroom rental listing in this large city is selected at random. What can be said about the probability that the listed rent price will be at least $930?

Nothing, beause the distribution of the monthly rental prices are not normal.

(b) Suppose a random sample 30 one-bedroom rental listing in this large city will be selected, the rent price will be recorded for each listing, and the sample mean rent price will be computed. What can be said about the probability that the sample mean rent price will be greater than $900?

Now we can apply the Central Limit Theorem.

\mu = 880, \sigma = 50, n = 30, s = \frac{50}{\sqrt{30}} = 9.1287

This probability is 1 subtracted by the pvalue of Z when X = 900.

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

By the Central Limit Theorem

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

Z = \frac{900 - 880}{9.1287}

Z = 2.19

Z = 2.19 has a pvalue of 0.9857

1 - 0.9857 = 0.0143

1.43% probability that the sample mean rent price will be greater than $900

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Answer: The total number of different types of labels will the manufacturers have to produce = 24.

Step-by-step explanation:

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Total number of different types of labels will have to produce = (Choices for flavors) x (Choices for sugar)x (Choices for the quantity )

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Answer:

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Step-by-step explanation:

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Let's start by solving the first equation.

a) -3 = 7 + 2t/3 
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Next, we should subtract 7 from both sides to cancel out the 7 on the right side.
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Finally, we should divide both sides by 2.
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Now let's move on to the next equation.

b) 4(5x-2) = 7(2x+3)
Let's use the distributive property to get rid of the parentheses and their coefficients.
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Now, lets subtract 14x from both sides of the equation.

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Next, let's add 8 to both sides of the equation.

6x = 29

And divide both sides by the coefficient of x, which is 6.

x = 29/6 or 4 5/6

Now for the last equation.

C) 2x - 6 = 20 - 2.5x
First, we should add 2.5x to both sides to cancel out the -2.5x on the right side of the equation.
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The probability of a biased coin landing on heads is 2/7 complete the tree diagram
mamaluj [8]

Answer:

The numbers needed to fill each box are in the image attached

Step-by-step explanation:

The probability of the coin landing on heads is 2/7, so the probability of it landing on tails is 1 - 2/7 = 5/7

The probability of landing heads 2 times is:

P = (2/7) * (2/7) = 4/49

The probability of landing heads and then tails is:

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The probability of landing tails and then heads is:

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The probability of landing tails 2 times is:

P = (5/7) * (5/7) = 25/49

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That would be zero.


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