1/2. It isn't asking for another solution, just the fraction.
Hope this helps!
Answer:
The maximum value of the confidence interval for this set of survey results is 51.73%.
Step-by-step explanation:
A confidence interval has two bounds, a lower bound and an upper bound.
These bounds depend on the sample proportion and on the margin of error.
The lower bound is the sample proportion subtracted by the margin of error.
The upper bound is the margin of error added to the sample proportion.
In this question:
Sample proportion: 46.1%
Margin of error: 5.63%.
Maximum value is the upper bound:
46.1+5.63 = 51.73
The maximum value of the confidence interval for this set of survey results is 51.73%.
Answer:
95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm)
Step-by-step explanation:
Our sample size is 11.
The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So
.
Then, we need to subtract one by the confidence level 
and divide by 2. So:
Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 10 and 0.025 in the two-sided t-distribution table, we have 
Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So
Now, we multiply T and s
![M = Ts = 1.812*1.2060 = 2.19/tex]For the lower end of the interval, we subtract the sample mean by M. So the lower end of the interval here is[tex]L = 97 - 2.19 = 94.81 = 94.8](https://tex.z-dn.net/?f=M%20%3D%20Ts%20%3D%201.812%2A1.2060%20%3D%202.19%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EFor%20the%20lower%20end%20of%20the%20interval%2C%20we%20subtract%20the%20sample%20mean%20by%20M.%3C%2Fstrong%3E%20So%20the%20lower%20end%20of%20the%20interval%20here%20is%3C%2Fp%3E%3Cp%3E%5Btex%5DL%20%3D%2097%20-%202.19%20%3D%2094.81%20%3D%2094.8)
cm
For the upper end of the interval, we add the sample mean and M. So the upper end of the interval here is
cm
So
95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm).
Answer:
P = 0.05
Step-by-step explanation:
12 months * 30 days each = 360 days
From 306 days, we have to select 8 days = 360C8 ways(Total ways)
We want each days from different month. First, we have to select 8 month from 12 month = 12C8 ways
---By selecting 8 month, we will select a days from each month. That can be done in = 30C1 * 30C1 * .................30C1 (8 ways) [From a month with 30 days, we can select a day in 30C1 ways = 30 ways]
Therefore P = Number of ways of selecting each days from different month / Total number of ways
P = 12C8 * 30^8 / 360C8
P = 495 * 656100000000 / 6469697679132645
P = 0.0501985588982791
P = 0.05
Hence the probability that each day is from a different month is 0.05
