Question 1193353: Please help me!
In the following exercises, use this information for the overhead reach distances of adult females: mean = 205.5 cm, S.D = 8.6 cm, and overhead reach distances are normally distributed. The overhead reach distances are used in planning assembly workstations.
1. If 1 adult female is randomly selected, find the probability that her overhead reach is less than 222.7 cm.
2. If 49 adult females are randomly selected, find the probability that they have a mean overhead reach of less than 207 cm.
3. If 1 adult female is randomly selected, find the probability that her overhead reach is greater than 196.9 cm.
4. If 36 adult females are randomly selected, find the probability that they have a mean overhead reach greater than 205 cm.
5. If 1 adult female is randomly selected, find the probability that her overhead reach is between 179.7 cm and 231.3 cm.
6. If 40 adult females are randomly selected, find the probability that they have a mean overhead reach between 204.0 cm and 206.0 cm.
Thank you for your time.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Problem 1
mu = population mean = 205.5
sigma = population standard deviation = 8.6
Convert the raw score x = 222.7 to its corresponding z score
z = (x - mu)/sigma
z = (222.7 - 205.5)/8.6
z = 2
Now use a Z table such as this one
https://www.ztable.net/
to find that P(Z < 2.00) = 0.97725
You'll need to scroll down to refer to the second table on that page.
Look at the row that starts with +2.0 and the column that starts with 0 up top. The row and column intersect at 0.97725
If you were to use a TI calculator, then you'd use the normalcdf( function. To access it, hit the yellow 2ND key at the top corner, then hit VARS to reach the stats distribution menu.
The general template of the function is
normalcdf(lower, upper)
so we must provide a lower and upper boundary under the curve (ie the left and right endpoints).
The right endpoint must be z = 2. The left endpoint can be some fairly large negative value such as -99
The screenshot shows that the area to the left of z = 2 is roughly 0.977249938; rounding to five decimal places gets us 0.97725 mentioned in the previous paragraph.
If you don't have a calculator, then you can use this free resource.
https://davidmlane.com/normal.html
Keep mean and SD as 0 and 1 respectively
Click the "below" button and type 2 in the box just next to it. Then hit "recalculate" to have 0.9772 show up.
If you use WolframAlpha (another free calculator), you would type in P(Z < 2) to have the result 0.97725 show up
I'll round the result to four decimal places, though be sure to follow your teacher's instructions.
Answer: Roughly 0.9773
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Problem 2
The sample size is n = 49
When forming the xbar distribution, it has a standard error (SE) of
SE = sigma/sqrt(n)
SE = 8.6/sqrt(49)
SE = 1.22857142857142
This will be treated as the standard deviation of the xbar distribution. The mean is the same as before
Let's convert the raw score x = 207 to its corresponding z score.
z = (x - mu)/SE
z = (207 - 205.5)/1.22857142857142
z = 1.22093023255814
z = 1.22
When the option arises, it's standard practice to round the z values to two decimal places. Though be sure to follow your teacher's instruction if s/he wants a different level of precision.
If you were to use the table I mentioned earlier, then you should get P(Z < 1.22) = 0.88877
Using a TI calculator should produce the value 0.8887674991
Using the calculator in the link I posted earlier would give the value 0.8888
If you prefer WolframAlpha, then type in P(Z < 1.22) to get a result of 0.888768
Answer: Approximately 0.8888
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Problem 3
We will use the formula mentioned in problem 1 to convert x = 196.9 to its z score
So,
z = (x - mu)/sigma
z = (196.9 - 205.5)/8.6
z = -1
Then use your method of choice (table or calculator) to find that P(Z < -1) is equal to...
0.15866 when using the table
0.1586552596 when using a TI calculator
0.1587 when using the David M Lane calculator link
0.158655 when using WolframAlpha
They all basically say that P(Z < -1) is equal to 0.1587 when rounding to four decimal places.
Then we can say this:
P(Z > -1) = 1 - P(Z < -1)
P(Z > -1) = 1 - 0.1587
P(Z > -1) = 0.8413
This converts to
P(X > 196.9) = 0.8413
when mu = 205.5 and sigma = 8.6
Answer: 0.8413 (approximate)
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Problem 4
Like with problem 2, we'll need to calculate the standard error (SE) first. This is because we're turning to the xbar distribution. Whenever we're asked to find the probability that the sample mean is larger than some cutoff, it's when we use the xbar distribution.
SE = sigma/sqrt(n)
SE = 8.6/sqrt(36)
SE = 1.43333333333333
Then we'd say
z = (x - mu)/SE
z = (205 - 205.5)/1.43333333333333
z = -0.34883720930232
z = -0.35
Using whatever method of choice (table or calculator), you should find that P(Z < -0.35) = 0.3632 when rounding to four decimal places
Therefore,
P(Z > -0.35) = 1 - P(Z < -0.35)
P(Z > -0.35) = 1 - 0.3632
P(Z > -0.35) = 0.6368
which converts over to
P(xbar > 205) = 0.6368
There's roughly a 63.68% chance we have a sample mean (xbar) larger than 205
Answer: 0.6368 (approximate)
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Problem 5
Convert x = 179.7 to its z score
z = (x-mu)/sigma
z = (179.7-205.5)/8.6
z = -3
Do the same for x = 231.3
z = (x - mu)/sigma
z = (231.3 - 205.5)/8.6
z = 3
Determining P(179.7 < X < 231.3) is exactly equivalent to P(-3 < Z < 3)
Use your method of choice to find that
P(Z < -3) = 0.0014
P(Z < 3) = 0.9987
So,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-3 < Z < 3) = P(Z < 3) - P(Z < -3)
P(-3 < Z < 3) = 0.9987 - 0.0014
P(-3 < Z < 3) = 0.9973
The nice thing about the WolframAlpha calculator is that you can type in P(-3 < Z < 3) as is, and it will produce roughly 0.9973 as a way to check your work.
The David M Lane calculator link also provides a direct route to finding the area between two given endpoints. Click on the "between" option.
With the TI calculator, you'll of course have the "lower" and "upper" as -3 and 3 respectively.
Answer: 0.9973 (approximate)
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Problem 6
sample size = n = 40
SE = standard error
SE = sigma/sqrt(n)
SE = 8.6/sqrt(40)
SE = 1.3597793938724
Convert xbar = 204.0 to its z score
z = (xbar - mu)/SE
z = (204.0 - 205.5)/1.3597793938724
z = -1.10312011401222
z = -1.10
Repeat for xbar = 206.0
z = (xbar - mu)/SE
z = (206.0 - 205.5)/1.3597793938724
z = 0.36770670467074
z = 0.37
Then we can say
P(Z < -1.10) = 0.1357
P(Z < 0.37) = 0.6443
so,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-1.10 < Z < 0.37) = P(Z < 0.37) - P(Z < -1.10)
P(-1.10 < Z < 0.37) = 0.6443 - 0.1357
P(-1.10 < Z < 0.37) = 0.5086
Answer: 0.5086
As with all of the other answers, this result is approximate to four decimal places.
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