Answer:
a. 0.76
b. 0.23
c. 0.5
d. p(B/A) is the probability that given that a student has a visa card, they also have a master card
p(A/B) is the probability that given a student has a master card, they also have a visa card
e. 0.35
f. 0.31
Step-by-step explanation:
a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)
=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76
b. P(AnBnC')= P(AnB)-P(AnBnC)
=0.3-0.07= 0.23
c. P(B/A)= P(AnB)/P(A)
=0.3/O.6= 0.5
e. P((AnB)/C))= P((AnB)nC)/P(C)
=P(AnBnC)/P(C)
=0.07/0.2= 0.35
f. P((AUB)/C)= P((AUB)nC)/P(C)
=(P(AnC) U P(BnC))/P(C)
=(0.11+0.1)/0.2
=0.21/0.2 = 0.31
Answer: 2
Step-by-step explanation: (2x-4)
Part. %
------ = -------
Whole 100
816. X
------- = ------
850. 100
850x=81600
-------- ---------
850. 850
X=96
96% of seats were sold
You can because you already know the number is 3 and the blank number is n
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:
The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):
Finally, the interval at 98% confidence level is:
