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BigorU [14]
4 years ago
8

In the casino game of roulette, a gambler can bet on which of 38 numbers the ball will land when the roulette wheel is spun. On

a $2 bet, a gambler gains $70 (so a net profit of $68) if he or she chooses the winning number, but loses the $2 otherwise. The probability of choosing the winning number is 1/38 and the probability of not choosing the winning number is 37/38. Let X denote the net gain from a roulette spin. What is the expected winnings for one spin?
Mathematics
1 answer:
ollegr [7]4 years ago
4 0

Answer:

Expected winnings for one spin = 0.1

Step-by-step explanation:

Expected Value = Sum of [(Probability of X)(Value of X)]

E(X) = Σ [P(X).X]

Two Events : X = +68 , P(X) = 1/38 and X = -2 , P(X) = 37/38

E(X) = [(68) (1/38)] + [(-2)(37/38)]

[(68)(0.03)] + [(-2)(0.97)]

= 2.04 - 1.94

= 0.1

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5 0
3 years ago
X=-4 +6+ (11 +4(-3))
lapo4ka [179]

Answer:

x=1

Step-by-step explanation:

x = -4 + 6 + (11 + 4(-3)).

First get the numbers in the parentheses

x = -4 + 6 + (11 - 12)

Since 11 - 12 = -1. We can replace (11-12) with -1

x = -4 + 6 -1

Now add and subtract

x = 1

4 0
3 years ago
Suppose that administrators at a large urban high school want to gain a better understanding of the prevalence of bullying withi
boyakko [2]

Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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".  

Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

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

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

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

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

5 0
4 years ago
Find the median first quarrile, third quartile, interquartile range, and an outliers for each set of data
Igoryamba
The outlier is 68

The median is 102
First quartile: 87
Third Quartile: 115
The interquartile range is 87-115
3 0
3 years ago
Find a solution to the ineqality x &gt; 6.<br> A. 4 <br> B. 11<br> C. 2<br> D. 6
Delicious77 [7]
It would be B. 11 is bigger then 6
5 0
3 years ago
Read 2 more answers
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