Question
In: Statistics and Probability
The mean amount of life insurance per household is $121,000. This distribution is positively skewed. The...
The mean amount of life insurance per household is $121,000. This distribution is positively skewed. The standard deviation of the population is $35,000. Use Appendix B.1 for the z-values.
a. A random sample of 55 households revealed a mean of $124,000. What is the standard error of the mean? (Round the final answer to 2 decimal places.)
Standard error of the mean
b. Suppose that you selected 124,000 samples of households. What is the expected shape of the distribution of the sample mean?
Shape (Click to select)UniformNot normal, the standard deviation is unknownNormalUnknown
c. What is the likelihood of selecting a sample with a mean of at least $124,000? (Round the final answer to 4 decimal places.)
Probability
d. What is the likelihood of selecting a sample with a mean of more than $109,000? (Round the final answer to 4 decimal places.)
Probability
e. Find the likelihood of selecting a sample with a mean of more than $109,000 but less than $124,000. (Round the final answer to 4 decimal places.)
Probability
Solutions
orchestra answered 2 years agoRelated Solutions
The mean amount of life insurance per household is $122,000. This distribution is positively skewed. The...
The mean amount of life insurance per household is $122,000.
This distribution is positively skewed. The standard deviation of
the population is $42,000. Use Appendix B.1 for the
z-values.
a. A random sample of 60 households revealed a
mean of $127,000. What is the standard error of the mean?
(Round the final answer to 2 decimal places.)
Standard error of the mean ?
b. Suppose that you selected 127,000 samples of
households. What is the expected shape of the distribution...
The mean amount of life insurance per household is $111,000. This distribution is positively skewed. The...
The mean amount of life insurance per household is $111,000.
This distribution is positively skewed. The standard deviation of
the population is $39,000. Use Appendix B.1 for the
z-values.
a. A random sample of 40 households revealed a
mean of $115,000. What is the standard error of the mean?
(Round the final answer to 2 decimal places.)
Standard error of the mean
b. Suppose that you selected 115,000 samples of
households. What is the expected shape of the...
The mean amount of life insurance per household is $110,000. This distribution is positively skewed. The...
The mean amount of life insurance per household is $110,000.
This distribution is positively skewed. The standard deviation of
the population is $40,000. Use Appendix B.1 for the
z-values.
a. A random sample of 50 households revealed a
mean of $112,000. What is the standard error of the mean?
(Round the final answer to 2 decimal places.)
Standard error of the mean
b. Suppose that you selected 50 samples of
households. What is the expected shape of the...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $115,000. This distribution follows the normal
distribution with a standard deviation of $37,000.
What is the likelihood of selecting a sample with a mean of at
least $120,000? (Round your z value to 2 decimal
places and final answer to 4 decimal places.)
What is the likelihood of selecting a sample with a mean of more
than $106,000? (Round...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $126,000. This distribution follows the normal
distribution with a standard deviation of $40,000.
If we select a random sample of 64 households, what is the
standard error of the mean? (Round your answer to the
nearest whole number.)
What is the expected shape of the distribution of the sample
mean?
What is the likelihood of selecting a sample with...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $110,000 with a standard deviation of $40,000. Assume the
population distribution is normal. A random sample of 100
households is taken.
What is the probability that sample mean will be less than
$120,000?
What is the probability that sample mean will be between
$105,000 and $120,000?
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $128,000. This distribution follows the normal
distribution with a standard deviation of $33,000.
If we select a random sample of 66 households, what is the
standard error of the mean? (Round your answer to the
nearest whole number.)
What is the expected shape of the distribution of the sample
mean?
What is the likelihood of selecting a sample with...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $132,000. This distribution follows the normal
distribution with a standard deviation of $40,000.
If we select a random sample of 70 households, what is the
standard error of the mean? (Round your answer to the
nearest whole number.)
What is the expected shape of the distribution of the sample
mean?
What is the likelihood of selecting a sample with...
Information from the American Institute of Insurance indicates the mean amount of life insurance per household...
Information from the American Institute of Insurance indicates
the mean amount of life insurance per household in the United
States is $134,000. This distribution follows the normal
distribution with a standard deviation of $30,000.
What is the likelihood of selecting a sample with a mean of at
least $137,000? (Round your z value to 2 decimal
places and final answer to 4 decimal places.)
d. What is the likelihood of selecting a sample with a mean of
more than $126,000?...
Information from the American Institute of Insurance indicates that the mean amount of life insurance per...
Information from the American Institute of Insurance indicates
that the mean amount of life insurance per household in the United
States is $115,000 and standard deviation is $41,250.
a) If a random sample of 32 households are selected, what is the
shape of the sampling distribution of the sample means?
b) Specify the mean and standard deviation of the sampling
distribution of the sample means, for samples of size 32.
c) What is the probability that in a random sample...
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