Complete question:
The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line
Answer:
(3.699, 4.701)
Step-by-step explanation:
Given:
Sample size, n = 45
Sample mean, x' = 4.2
Standard deviation
= 2.0
Required:
Find a 90% CI for true mean time
First find standard error using the formula:




Standard error = 0.298
Degrees of freedom, df = n - 1 = 45 - 1 = 44
To find t at 90% CI,df = 44:
Level of Significance α= 100% - 90% = 10% = 0.10

Find margin of error using the formula:
M.E = S.E * t
M.E = 0.298 * 1.6802
M.E = 0.500938 ≈ 0.5009
Margin of error = 0.5009
Thus, 90% CI = sample mean ± Margin of error
Lower limit = 4.2 - 0.5009 = 3.699
Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701
Confidence Interval = (3.699, 4.701)
We can use the sum of an aritmetic sequence
the sum from n=1 to n=r when the first term is a1 and the nth term is an is

first term is 1
last term is 100
there are 100 terms so n=100
so the sum is


S=5050
now you want us to divide by 10
5050/10=505
fun fact, gauss (famous math guy) did this when he was younger, legend has it that he was assigned this as an in class assigment to kill time but gauss found a neat pattern, he noticed that adding the end terms wer giving the same sum, example, 100+1=101, 2+99=101, etc, so he just needed to find al the pairs and add them all up
answer is 505
the result is 505
Answer:
x therefore will be equal to 4
Huh this gives out no information to answer