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Answer: If we imagine a normal curve and draw a line at the $33,000 point
and another line at the $34,000 point (which happens to be the population
mean) we can then visually see the area where question 10a is referring.
What percentage of cases in our sample is expected to fall in this area
(between $33,000 and the mean of $34,000). That number will be the
probability that our sample mean would fall in this area (the answer to 10a).
So, since the mean is $34,000, all we have to do is determine the Z score
for $33,000, look at the table of Z scores and examine the percentage
score found under column B which is the area between Z and the mean.
What makes this problem more difficult is that we need to use a slightly
different formula in order to calculate the Z score since we have
information about the population that we can use. The slightly different
formula is the second formula provided in the learning check.
Substituting this formula for the formula above it (used when we have a
sample but don’t have information about the whole population) would be easy
to do except that you are only just learning the terminology. Consequently,
this slightly different formula uses terms that you are less familiar with
which makes it difficult to figure out how to plug numbers into the formula.
So, the learning check states “However, because here we are dealing with a
sampling distribution” (actually a sample of 200 that we will treat as if it
were a sampling distribution) “replace Y with the sample mean (or $33,000)
and replace ‘Y mean’ with the sampling distribution’s mean (actually the
population’s mean since we know that a sampling distribution’s mean is
equivalent to the population’s mean and we have the population mean
available), and the standard deviation with the standard error of the mean.”
So first we may want to calculate the standard error of the mean. This is
accomplished by taking the square root of the sample N (200) and then
dividing it into the population’s standard deviation (the formula is provided in
the learning check—it is found in the bottom formula, and within this
formula is the bottom half).
That is, the standard error of the mean = $5,000/ sqrt of 200 = $353.55
By fitting the SE ($353.55) into the whole formula we can then obtain the
Z score for a sample mean of $33,000.
Z = $33,000 - $34,000/ 353.55 = -2.83
The area between the Z score of 2.83 (converted from $33,000) and the
mean ($34,000) is about .4977 which is also the probability of a mean
between $33,000 and $34,000. That is, there is a 49.8% probability that
our sample mean would fall between $33,000 and $34,000.
To take this a step beyond the question being asked, what would be the
probability that our sample mean would fall between $35,000 and the mean
($34,000)? This is really easier than it looks. Since $33,000 is $1,000 less
than the mean and $35,000 is $1,000 more than the mean, their percentages
will be the same (.4977) with $33,000 drawn to the right of the mean
(34,000) and $35,000 drawn to the left of the mean.
Normal distribution z score is 0.4977
Normal distribution z score based problem;
Given information in question:
Here mean μ = 34,000
Standard deviation σ = 5000
Sample size n = 200
Find:
Normal distribution z score
Computation:
Standard error = σ/√n = 5000 / √200
Standard error = 353.5534
Probability = P(33000 < X < 34000)
Probability = P((33000-34000)/353.553) < Z < (34000-34000)/353.553)
Probability =P(-2.83 < Z < 0)
Probability =0.5-0.0023
Probability =0.4977
Find out more information about Standard error'.
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