For the total population of a large southern city mean family is income 34,000 with a standard deviation for the population of 5000 imagine that you were taking some sample of 200 city residents what is the probability or proportion that your sample mean is between 33,000 and 34000

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

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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