Answer:
Step-by-step explanation:
Given that a researcher is trying to decide how many people to survey.
We have confidence intervals are intervals with middle value as the mean and on either side margin of error.
Confidence interval = Mean ± Margin of error
Thus confidence interval width depends on margin of error.
Margin of error = 
Thus for the same confidence level and std deviation we find margin of error is inversely proportional to square root of sample size.
Hence for small n we get wide intervals.
So if sample size = 300, the researcher will get wider confidence interval
Divide the paper which is 21 inches (12+9) by 1.5 and you have your answer which is 14
Step-by-step explanation:
Answer: 0.1357
Step-by-step explanation:
Given : Monitors manufactured by TSI Electronics have life spans that have a normal distribution with a variance of 
and a mean life span of 
hours.
Here , 
Let x represents the life span of a monitor.
Then , the probability that the life span of the monitor will be more than 14,650 hours will be :-
![P(x>14650)=P(\dfrac{x-\mu}{\sigma}>\dfrac{14650-13000}{1500})\\\\=P(z>1.1)=1-P(z\leq1.1)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\=1-0.8643339=0.1356661\approx0.1357](https://tex.z-dn.net/?f=P%28x%3E14650%29%3DP%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B14650-13000%7D%7B1500%7D%29%5C%5C%5C%5C%3DP%28z%3E1.1%29%3D1-P%28z%5Cleq1.1%29%5C%20%5C%20%5B%5Cbecause%5C%20P%28Z%3Ez%29%3D1-P%28Z%5Cleq%20z%29%5D%5C%5C%5C%5C%3D1-0.8643339%3D0.1356661%5Capprox0.1357)
Hence, the probability that the life span of the monitor will be more than 14,650 hours = 0.1357
The answer is 8788pi/3 Hope this helps! Mark brainly please!
