Aproduct comes in cans labeled "38 oz". a random sample of 10 cans had the following weights:

{34.6, 39.65, 34.75, 40, 39.5, 38.9, 34.25, 36.8, 39, 37.2} with the following summary statistics:

n equals 10 space space top enclose x equals 37.465 space space s equals 2.27

estimate the true mean weight of the cans of product with 95% confidence.

a. 34, 41 ounces
b. 35.9, 39.1 ounces
c. 36.1, 38.9 ounces
d. 37.465 ounces

b. 35.9, 39.1 ounces

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

barX=37.465 represent the sample mean for the sample  

population mean (variable of interest)

s=2.27 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

barX +- t*[(s)/(sqrt{n))]   (1)

In order to calculate the critical value t we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_(alpha/2)=2.26

Now we have everything in order to replace into formula (1):

37.465-2.26[(2.27)/(sqrt{10))]=35.9    

37.465+2.26[(2.27)/(sqrt{10))]=39.1

So on this case the 95% confidence interval would be given by (35.9;39.1)    

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %