Use the diagram to solve x and y
1 answer:
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Answer:
Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .
Step-by-step explanation:
We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.
Also, a sample of four metal sheets is randomly selected from a batch.
Let X bar = Average length of a sheet
The z score probability distribution for average length is given by;
Z = ~ N(0,1)
where, = population mean = 30.05 inches
= standard deviation = 0.2 inches
n = sample of sheets = 4
So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)
P(30.25 inches < X bar < 30.35 inches) = P(X bar < 30.35) - P(X bar <= 30.25)
P(X bar < 30.35) = P( <
) = P(Z < 3) = 0.99865
P(X bar <= 30.25) = P( <=
) = P(Z <= 2) = 0.97725
Therefore, P(30.25 inches < X bar < 30.35 inches) = 0.99865 - 0.97725
= 0.0214
Answer:
The 80% confidence interval for the mean consumption of meat among people over age 23 is between 4 and 4.2 pounds.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the mean subtracted by M. So it is 4.1 - 0.07 = 4.03 pounds
The upper end of the interval is the mean added to M. So it is 4.1 + 0.07 = 4.17 pounds
Rounded to one decimal place
The 80% confidence interval for the mean consumption of meat among people over age 23 is between 4 and 4.2 pounds.