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4 years ago

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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)?

1 answer:

snow_tiger [21]4 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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Complete Question

The complete question is shown on the first uploaded image

Answer:

the null hypothesis is $H_o : \mu = 122$

the alternative hypothesis is $H_a : \mu \ne 122$

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Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 122$

The sample size is n= 38

The sample mean is $\= x = 116 \ feet$

The standard deviation is $\sigma = 21$

Generally the null hypothesis is $H_o : \mu = 122$

the alternative hypothesis is $H_a : \mu \ne 122$

Generally the test statistics is mathematically evaluated as

$t = \frac { \= x - \mu }{\frac{ \sigma }{ \sqrt{n} } }$

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$t = \frac { 116 - 122 }{\frac{ 21 }{ \sqrt{ 38} } }$

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The p-value is mathematically represented as

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$p = P(Z < t ) = 0.039119$

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