Answer:
Expected pay winning $50= $0.585
Expected pay winning $25= $2.36
Expected pay for anything else= $-4.35
Expected returns=3.59
Expected value for one play= $(-1.41)
Do not play this game because you will lose $1.41
Step-by-step explanation:
Probability P(3 hearts) = (13/52)×(12/51)×(11/50) = 0.013
Probability P(3black)= (26/52)×(24/51)×(23/50) = 0.118
Probability P(drawing anything else)= 1 - 0.013 - 0.118= 0.869
Expected pay($50)= 0.013$(50-5)= $ 0.585
Expected pay($25)= 0.118(25-5)$ = $2.36
Expected pay for anything else= 0.869(0-5)$ =$(-4.347)
Expected value of one play=$ (0.585 + 2.353 -4.347) = -$1.41
c) Do not play the game.
5d + 2(2 - d) = 3(1 + d) + 1
5d + 2(2) - 2(d) = 3(1) + 3(d) + 1
5d + 4 - 2d = 3 + 3d + 1
5d - 2d + 4 = 3d + 3 + 1
3d + 4 = 3d + 4
<u>-3d -3d </u>
4 = 4
d = 0
8 total pens....4 are black
first pick, probability of being black is 4/8
2nd pick. without replacing, probability is 3/7
probability of a black pen picked first and then another black pen picked again is : 4/8 * 3/7 = 12/56 = 0.21
Answer:
save up and pay cash
Step-by-step explanation:
paying cash is always the best idea and not trying to get so much debit. If you did one other otion and have the debit some emergency can come up and you will not be able to pay for that, and aying cash you save more money
Answer:
The 95% confidence interval for the population variance is ![\left[0.219, \hspace{0.1cm} 0.807\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B0.219%2C%20%5Chspace%7B0.1cm%7D%200.807%5Cright%5D%5C%5C%5C%5C)
The 95% confidence interval for the population mean is ![\left [15.112, \hspace{0.3cm}15.688\right]](https://tex.z-dn.net/?f=%5Cleft%20%5B15.112%2C%20%5Chspace%7B0.3cm%7D15.688%5Cright%5D)
Step-by-step explanation:
To solve this problem, a confidence interval of 
for the population variance will be calculated.
Then, the confidence interval for the population variance is given by:
![\left [\frac{(n-1)S^2}{\chi^2_{\left (\frac{\alpha}{2};n-1\right )}}, \hspace{0.3cm}\frac{(n-1)S^2}{\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}} \right ]\\\\](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%28n-1%29S%5E2%7D%7B%5Cchi%5E2_%7B%5Cleft%20%28%5Cfrac%7B%5Calpha%7D%7B2%7D%3Bn-1%5Cright%20%29%7D%7D%2C%20%5Chspace%7B0.3cm%7D%5Cfrac%7B%28n-1%29S%5E2%7D%7B%5Cchi%5E2_%7B%5Cleft%20%281-%5Cfrac%7B%5Calpha%7D%7B2%7D%3Bn-1%5Cright%20%29%7D%7D%20%5Cright%20%5D%5C%5C%5C%5C)
Thus, the 95% confidence interval for the population variance is:![\\\\\left [\frac{(19-1)(0.6152)^2}{32.852}, \hspace{0.1cm}\frac{(19-1)(0.6152)^2}{8.907} \right ]=\left[0.219, \hspace{0.1cm} 0.807\right]\\\\](https://tex.z-dn.net/?f=%5C%5C%5C%5C%5Cleft%20%5B%5Cfrac%7B%2819-1%29%280.6152%29%5E2%7D%7B32.852%7D%2C%20%5Chspace%7B0.1cm%7D%5Cfrac%7B%2819-1%29%280.6152%29%5E2%7D%7B8.907%7D%20%5Cright%20%5D%3D%5Cleft%5B0.219%2C%20%5Chspace%7B0.1cm%7D%200.807%5Cright%5D%5C%5C%5C%5C)
On other hand,
A confidence interval of for the population mean will be calculated
\![\left[ \bar X - T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}}, \hspace{0.3cm}\bar X + T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}} \right ]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbar%20X%20-%20T_%7B%28%5Calpha%2F2%3Bn-1%7D%5Csqrt%7B%5Cfrac%7B%5CS%5E2%7D%7Bn%7D%7D%2C%20%5Chspace%7B0.3cm%7D%5Cbar%20X%20%2B%20T_%7B%28%5Calpha%2F2%3Bn-1%7D%5Csqrt%7B%5Cfrac%7B%5CS%5E2%7D%7Bn%7D%7D%20%5Cright%20%5D%5C%5C%5C%5C)
Thus, the 95\% confidence interval for the population mean is:![\\\\\left [15.40 - 2.093\sqrt{\frac{(0.6152)^2}{19}}, \hspace{0.3cm}15.40 + 2.093\sqrt{\frac{(0.6152)^2}{19}} \right ]=\left [15.112, \hspace{0.3cm}15.688\right] \\\\](https://tex.z-dn.net/?f=%5C%5C%5C%5C%5Cleft%20%5B15.40%20-%202.093%5Csqrt%7B%5Cfrac%7B%280.6152%29%5E2%7D%7B19%7D%7D%2C%20%5Chspace%7B0.3cm%7D15.40%20%2B%202.093%5Csqrt%7B%5Cfrac%7B%280.6152%29%5E2%7D%7B19%7D%7D%20%5Cright%20%5D%3D%5Cleft%20%5B15.112%2C%20%5Chspace%7B0.3cm%7D15.688%5Cright%5D%20%5C%5C%5C%5C)
