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Olenka [21]
3 years ago
9

Find the expected value of the random variable. Round to the nearest cent unless stated otherwise. Suppose you buy 1 ticket for

$1 out of a lottery of 1,000 tickets where the prize for the one winning ticket is to be $500. What is your expected value
Mathematics
1 answer:
QveST [7]3 years ago
5 0

Answer:

Expected Value = - 0.5

Step-by-step explanation:

Expected Value [ E(X) ]  = Σ  {X. Probability (X)} = Σ [ (X).P(X) ]

  • Case 1 : Lottery Won

Probability = 1 / 1000

X Value = 500 - 1 = 499

  • Case 2 : Lottery Lost

Probability =  1 - Probability Winning = 999 / 1000

X Value = -1

  • Expected Value :

(1 / 1000) (499) + (999 / 1000) (-1)

0.499 - 0.999

= - 0.5

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the
AVprozaik [17]

Answer:

a) n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

b) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

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{\hat p(1-\hat p)}{n}})  

The margin of error for the proportion interval is given by this formula:  

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

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

Part a

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 90% of confidence, our significance level would be given by \alpha=1-0.9=0.1 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.64  

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.02 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

Part b

For this case since we don't have a prior estimate we can use \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

8 0
3 years ago
help please <br><br> No links
11Alexandr11 [23.1K]

Answer:

62-0.75x8

Step-by-step explanation

$6=discount

8 tickets

Total=62.00

0.75 multiplied by 8= 6

62-6=56

56 is before the discount

Equation would be 62-0.75x8

8 0
2 years ago
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