X
1.
A sample of size n=400 is randomly selected from a non-normal population
with a mean =240 and standard deviation =40.
a.
Find P( X  235) .9938
b.
Find P( 236  X  244) .9544
c.
Find P( X  245) .9938
2.
At a certain bank, the mean checking account balance is =$950 and the
population standard deviation is =$104. (Assume the distribution is
normal).
a. What is the probability that the sample mean of 256 randomly chosen
checking account balances is between $944 and $956? .6424
b. What is the probability that a randomly selected individual checking
account is between $944 and $956? .0478
c. What is the probability that the sample mean of 256 randomly chosen
checking account balances is between $940 and $960? .8764
d. What is the probability that a randomly selected individual checking
account is between $940 and $960? .0796
3.
Assume the average city in the US has 45,200 residents and the population
standard deviation is 33,800. Suppose you randomly sample 49 American
cities.
a. Describe the population you are sampling from. Do you think the
distribution of city sizes would be normal?
The population consists of US cities. The distribution is likely skewed
to the right due to larger cities such as New York and Los Angeles.
b. Calculate the probability the sample mean city size is greater than
52,000.
.0793
c. Calculate the probability the sample mean city size is less than 48,000.
.7190
d. Suppose that instead of randomly sampling across all American cities,
you purposely selected only among cities in California. How would this
change the probabilities calculated in parts b and c?
Your sampling suggests you are estimating the population mean city
size in California, not the mean city in the US. The sampling procedure
is biased if the defined population remains all US cities. Unless the
mean and variance of the distribution of California cities match that for
the US as a whole, the probabilities you would calculate using the
California sample would be incorrect.
e. Given the information we have, explain why we cannot calculate the
probability that an individual randomly selected city would be within a
certain size range.
Because the distribution is not normal. We don’t know what the
distribution of the population of city sizes is.
4.
In 2002, the population mean yearly tuition at state universities in the US
was $4,260. Assume the population standard deviation is $900.
a. Calculate the probability that a sample of 36 state universities has a
mean tuition within $200 of the population mean. .8164
b. Calculate the probability that a sample of 100 state universities has a
mean tuition within $200 of the population mean. .9736
c. Explain why the probability calculated in part b is larger than the
probability calculated in part a.
The larger sample decreases the standard deviation of X .
5.
Suppose the population mean number of books at non-academic libraries in
the US is 35,500 books with a population standard deviation of 15,000
books. Assume you randomly sample 64 libraries.
a. Describe the population of the problem.
All non-academic libraries in the US.
b. Calculate the probability of obtaining a sample mean number of books
greater than 40,000.
.0082
c. Calculate the probability of obtaining a sample mean number of books
greater than 33,625.
.8413
d. P(X > 50,000) ≈0. Verify that this is true and explain what it means.
Does this imply that there is no public library in the US with more than
50,000 books?
P(X >50,000)= P(Z >7.73)≈0
No, this is only saying that the chance that a sample mean is greater than
50,000 is approximately zero. The chance that an individual library has
over 50,000 books is P(X>50,000), which equals .166.
e. Explain why the population of library sizes (in terms of number of
books) is probably not normally distributed.
The distribution of library sizes probably has the same general shape as the
distribution of city sizes. The outlier cities on the right side of the
distribution probably have large central libraries that would be outliers in
the distribution of libraries.