Answer:
99% CI: [45.60; 58.00]min
Step-by-step explanation:
Hello!
Your study variable is:
X: Time a customer stays in a certain restaurant. (min)
X~N(μ; σ²)
The population standard distribution is σ= 17 min
Sample n= 50
Sample mean X[bar]= 51.8 min
Sample standard deviation S= 27.68
You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:
X[bar] ±
*(σ/√n)

Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min
Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min
With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.
I hope you have a SUPER day!
PS: Missing Data in the attached files.
Answer:
There are 8 numbers.
4 of the numbers are even ( 2, 4, 6, 8)
The probability is the number of even numbers over the total numbers:
4/8 which reduces to 1/2.
The answer is 1/2
Answer:
if your doing scientific notation then 2.7x10^3 would be the only one correct. if its not scientific notation then 27x10^3 would be more because it would be equivilent to <em>2700<27000</em>
Step-by-step explanation: