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mash [69]
3 years ago
13

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th

e highest bid in excess of $9,600 will be accepted. Assume that the competitor's bid x is a random variable that is uniformly distributed between $9,600 and $14,500. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?
Mathematics
1 answer:
snow_tiger [21]3 years ago
6 0

Answer:

0.49 = 49% probability that your bid will be accepted

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

P(X \leq x) = \frac{x - a}{b-a}

Uniformly distributed between $9,600 and $14,500.

This means that a = 9600, b = 14500

Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?

Your bid will be accepted if the competitor's bid X is lower than 12000. So

P(X \leq 12000) = \frac{12000 - 9600}{14500 - 9600} = 0.49

0.49 = 49% probability that your bid will be accepted

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Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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idk tbh ajajaanananaa aja an aja a

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