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AysviL [449]
3 years ago
9

In a certain card game you draw one card off a standard deck of 52 cards. If you draw a spade you get paid $12, if you draw a re

d Ace you get paid $20, and if you draw a red Queen you get paid $38. If you draw anything else, you get paid nothing. What should this game cost if it is to be a fair game? Use fractions in your work and then calculate the answer as a decimal rounded to 4 decimal places.
Mathematics
1 answer:
Mariulka [41]3 years ago
4 0

Step-by-step explanation:

In a standard deck of 52 cards, there are 2 red aces, 2 red Queens, and 13 spades.  That leaves 35 cards for everything else.

For the game to be fair, the cost must equal the expected value.  The expected value is the sum of each outcome times its probability.

C = (12) (13/52) + (20) (2/52) + (38) (2/52) + (0) (35/52)

C = 68/13

C ≈ 5.2308

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An inspector inspects large truckloads of potatoes to determine the proportion p in the shipment with major defects, prior to us
Sindrei [870]

Answer:

0.06 - 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.027

0.06 + 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.093

The 95% confidence interval would be given by (0.027;0.093)

A. 0.027 to 0.093.

Step-by-step explanation:

Notation and definitions

X=12 number of defective.

n=200 random sample taken

\hat p=\frac{12}{200}=0.06 estimated proportion of defectives

p true population proportion of defectives

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.06 - 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.027

0.06 + 1.96\sqrt{\frac{0.06(1-0.06)}{200}}=0.093

The 95% confidence interval would be given by (0.027;0.093)

A. 0.027 to 0.093.

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