Question 1205471
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Check out this article
https://www.statology.org/normal-approximation/


If you're still stuck then move to the next section shown below.


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Part (a)


n = sample size = 80
p = probability of selecting a woman = 0.20


Computing {{{np}}} and {{{n(1-p)}}} leads to 16 and 64 respectively.
Both results are larger than 5, so they meet the criteria {{{np >= 5}}} and {{{n(1-p) >= 5}}}
Some stats textbooks will use 10 in place of 5; luckily 16 and 64 meet that threshold as well.
We confirm that a normal approximation to the binomial can be used.
Make sure to do this check before going any further. 


mu = mean
sigma = standard deviation


mu = n*p
mu = 80*0.20
mu = 16


sigma = sqrt(n*p*(1-p))
sigma = sqrt(80*0.20*(1-0.20))
sigma = 3.577708764 approximately


Refer to the link I posted to account for the continuity correction. This is where we go from a discrete setting to a continuous setting.


x = 16 leads to the interval 15.5 < x < 16.5


Use any stats calculator you want to find the area under the normal distribution curve between those endpoints.
I'll use a TI84.


If you are using that as well then press the button labeled "2nd".
Then press the VARS key.
Move to <font color=green>normalCDF</font>


The full command to type in would be <font color=green>normalCDF(15.5, 16.5, 16, 3.577708764)</font>
The standard template is <font color=green>normalCDF(L, R, mu, sigma)</font>
L = left endpoint
R = right endpoint


The result of the calculation <font color=green>normalCDF(15.5, 16.5, 16, 3.577708764)</font> is <font color=red>approximately 0.11114589</font>
Round that value however needed.


Therefore, 
P(15.5 < x < 16.5) = <font color=red>0.11114589</font> approximately.


There's about an 11.1146% chance of having 16 women in the sample of 80 employees.


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Part (b)


"14 or more" translates to {{{x >= 14}}}


Revisit the continuity correction chart in this link
https://www.statology.org/normal-approximation/


Apply the continuity correction to make {{{x >= 14}}} convert to x > 13.5
We subtract off 0.5 from 14 to get this new endpoint.


Using the TI84 we could type in
<font color=green>normalCDF(13.5, 9999, 16, 3.577708764)</font>
Notice there technically isn't a right endpoint. 
The 9999 is some really large number to effectively stand in for positive infinity.


The TI84 calculator says the result is roughly <font color=red>0.75765257</font>


Another calculator you could use is this
https://davidmlane.com/normal.html
It offers a diagram to go along with the area value.


Another alternative is the <a href="https://wiki.geogebra.org/en/Normal_Command">Normal</a> command in GeoGebra. 
Or you could use the <a href="https://support.microsoft.com/en-us/office/normdist-function-126db625-c53e-4591-9a22-c9ff422d6d58">normDist</a> command in a spreadsheet.
There are many other calculators that provide the same functionality.


If your teacher wants you to use a Z table instead of a calculator, then be sure to follow those instructions. You'll need to convert to a z score first before using the table. 
 

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Part (c)


Use the continuity correction chart to go from {{{x <= 15}}} to x < 15.5


Whichever calculator or table you end up using, the answer should be approximately <font color=red>0.44442708</font>
The answer may slightly vary depending on what method you used.


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I'll leave the other parts for the student to do.
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