Answer:
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.
The probability of k online retail orders that turn out to be fraudulent in the sample is:
We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:
![P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483](https://tex.z-dn.net/?f=P%28x%5Cgeq2%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%5D%5C%5C%5C%5C%5C%5CP%28x%3D0%29%3D%5Cdbinom%7B20%7D%7B0%7D%5Ccdot0.08%5E%7B0%7D%5Ccdot0.92%5E%7B20%7D%3D1%5Ccdot1%5Ccdot0.189%3D0.189%5C%5C%5C%5C%5C%5CP%28x%3D1%29%3D%5Cdbinom%7B20%7D%7B1%7D%5Ccdot0.08%5E%7B1%7D%5Ccdot0.92%5E%7B19%7D%3D20%5Ccdot0.08%5Ccdot0.205%3D0.328%5C%5C%5C%5C%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-%5B0.189%2B0.328%5D%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-0.517%3D0.483)
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
It is given that:
Monthly Rent = A
Total Money Paid to Landlord = Y
The landlord requires initial payment to include:
- 1st month's rent
month's rent or 50% of monthly rent for Security Deposit- 40% of monthly rent for utility bills
- 5% of monthly rent for health club
- 13% of monthly rent per person for other services
Also, Total Amount Paid to the Landlord = $2574
Number of People Sharing the Apartment = 3
Basis the above information,
Y = (100% * A) + (50% * A) + (40% * A) + (5% * A) + (13% * A * 3)
⇒ Y = A + (0.5 * A) + (0.4 * A) + (0.05 * A) + (0.39 * A)
⇒ Y = A + (0.5 * A) + (0.4 * A) + (0.05 * A) + (0.39 * A)
⇒ Y = 2.34 * A
Substituting the value of Y to determine A
⇒ 2,574 = 2.34 * A
⇒ A = 1,100
Hence, the monthly rent is $1,100
Answer: The least score she can get on her next test is 455 points.
Step-by-step explanation: From the available information, her current mean score is 88%, and that was achieved from a total of 5 tests. Hence, her mean score was computed as follows;
Mean = (Summation of data)/observed data
Where the observed data is 5 and the mean is 88. Therefore the mean can now be expressed as,
88 = Summation of data/5
By cross multiplication we now have
88 x 5 = Summation of data
440 = Summation of data.
So, if she wants to raise her mean score on her next test to 91%, her least possible score would be derived as
91 = (Summation of data)/5
By cross multiplication we now have
91 x5 = Summation of data
455 = Summation of data
Therefore, on her next test, she must score at least 455 points.
Answer:
For this case the p value calculated is higher than the significance level used of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:
a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100
Step-by-step explanation:
Information given
We want to verify if he mean IQ of employees in an organization is greater than 100 , the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
The statistic for this case is given by:
(1)
The statistic calculated for this case 
The degrees of freedom are given by:
Now we can find the p value using tha laternative hypothesis and we got:
For this case the p value calculated is higher than the significance level used of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:
a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100
