QUESTION 1
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1/1 POINTS
Suppose germination periods, in days, for grass seed are normally distributed and have
a known population standard deviation of 2 days and an unknown population mean. A
random sample of 22 types of grass seed is taken and gives a sample mean
of 46 days. Find the error bound (EBM) of the confidence interval with
a 90% confidence level.
EBM=(zα2)(σn−−√)
We know that σ=2 and n=22. We are also given that the confidence level (CL)
is 90%, or 0.9. So, we can calculate alpha (α).
α=1−CL=1−0.9=0.1
Since α=0.1, we know that
α2=0.12=0.05
The value of z0.05 is 1.645. Now we can substitute the values into the formula to find
the error bound.
, EBM=(zα2)(σn−−√)=(1.645)(222−−√)≈(1.645)(0.426)≈0.701
So, the error bound (EBM) is 0.701.
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QUESTION 2
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1/1 POINTS
Suppose the number of pages per book in a library has an unknown distribution with
population mean 320 and population standard deviation 15. A sample of
size n=80 is randomly taken from the population, and the sum of the values is taken.
Using the Central Limit Theorem for Sums, what is the standard deviation for the
sample sum distribution? Round your answer to two decimal places.
The Central Limit Theorem for Sums states the standard deviation of the normal
distribution of sample sums is equal to the original distribution's standard deviation
multiplied by the square root of the sample size, (σX)(n−−√). The original standard
deviation is 15, and the sample size is 80. So, the standard deviation of the
distribution of sample sums is
(σX)(n−−√)=(15)(80−−√)≈134.16
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QUESTION 3
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1/1 POINTS
Suppose we know that a confidence interval is (61,71), with an error bound of 5.
Find the sample mean (x¯). Give just a number for your answer. For example, if you
found x¯=12, you would enter 12.
That is correct!
$$66
Answer Explanation
Correct answers:
$66$66
Since we know the error bound is 5, we can subtract 5 from the upper value of the
confidence interval, 71.
71−5=66
So the sample mean is 66; x¯=66.
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QUESTION 4
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