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Marrrta [24]
3 years ago
13

Suppose a basketball player has made 390 out of 434 free throws. If the player makes the next 2 free throws, I will pay you $40.

Otherwise you pay me $169
Step 1 of 2: Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.

Step 2 of 2: If you played this game 588 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be entered as negative.
Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0

Answer: a) -$0.19, b) -$111.72 .

Step-by-step explanation:

Since we have given that

Number of free throws = 434

Number of throws made by them = 390

Amount for making the next 2 free throws = $40

Amount otherwise he has to pay = $169

a) Find the expected value of the proposition.

Expected value of success in next 2 free throws = \dfrac{390}{434}\times \dfrac{391}{435}=0.8077

Expected value would be

0.8077\times 40+(1-0.8077)\times -169\\\\=32.308-32.4987\\\\=-\$0.19

b)  If you played this game 588 times how much would you expect to win or lose?

Number of times they played the game = 588

So, Expected value would be

588\times -0.19\\\\=-\$111.72

Hence, a) -$0.19, b) -$111.72

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Step-by-step explanation:

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Now, when we plug it in the equation, it looks like this:

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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

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Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

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