SOLUTION: Twenty percent of the employees of a large company are female. Use the normal approximation of the binomial probabilities to answer the following questions. What is the probabili

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Question 1205471: Twenty percent of the employees of a large company are female. Use the normal approximation
of the binomial probabilities to answer the following questions. What is the probability
that in a random sample of 80 employees
a. exactly 16 will be female?
b. 14 or more will be female?
c. 15 or fewer will be female?
d. 18 or more will be female
e. exactly 17 will be female?

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

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 and leads to 16 and 64 respectively.
Both results are larger than 5, so they meet the criteria and
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 normalCDF

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

The result of the calculation normalCDF(15.5, 16.5, 16, 3.577708764) is approximately 0.11114589
Round that value however needed.

Therefore,
P(15.5 < x < 16.5) = 0.11114589 approximately.

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

--------------------------------------------------

Part (b)

"14 or more" translates to

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

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

Using the TI84 we could type in
normalCDF(13.5, 9999, 16, 3.577708764)
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 0.75765257

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 Normal command in GeoGebra.
Or you could use the normDist 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 to x < 15.5

Whichever calculator or table you end up using, the answer should be approximately 0.44442708
The answer may slightly vary depending on what method you used.

--------------------------------------------------

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

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Twenty percent of the employees of a large company are female. Use the normal approximation
of the binomial probabilities to answer the following questions. What is the probability
that in a random sample of 80 employees
(a) exactly 16 will be female?
(b) 14 or more will be female?
(c) 15 or fewer will be female?
(d) 18 or more will be female
(e) exactly 17 will be female?
~~~~~~~~~~~~~~~~~~~~~~

You have a binomial distribution with large number of trials n = 80 
and individual probability of success p = 20% = 0.2.


You want to approximate it by the normal distribution.
You should use the mean value m = p*n = 0.2*80 = 16 and standard deviation 

    SD =  =  = 3.57771.


Also, you should use the continuity correction factor.


About approximation of the binomial distribution by normal distribution and continuity correction factor 
see your textbook and/or these Internet sources

https://www.statology.org/normal-approximation

https://www.statisticshowto.com/probability-and-statistics/binomial-theorem/normal-approximation-to-the-binomial/

https://online.stat.psu.edu/stat414/lesson/28/28.1

https://stats.libretexts.org/Courses/Las_Positas_College/Math_40%3A_Statistics_and_Probability/06%3A_Continuous_Random_Variables_and_the_Normal_Distribution/6.04%3A_Normal_Approximation_to_the_Binomial_Distribution


    For calculations, you may use your calculator (function normcdf), or Excel spreadsheet (function NORMDIST);
    or online calculator https://onlinestatbook.com/2/calculators/normal_dist.html



(a)  In this case, you should find the area under the normal curve between z-scores 15.5 and 16.5
                                                             (using the correction factor)

                          z1    z2     mean    SD                <<<---===  formatting pattern
     P(x = 16) = normcdf(15.5, 16.5,   16,    3.57771) = 0.1111  (rounded).



(b)  In this case, you should find the area under the normal curve on the right from 13.5
                                                             (using the correction factor)

                          z1    z2     mean    SD                <<<---===  formatting pattern
     P(x >= 14) = normcdf(13.5, 9999,  16     3.57771) = 0.7577  (rounded).   



(c)  In this case, you should calculate the area under the normal curve on the left of the mark z = 15.5
                                                                  (using the correction factor)

                           z1    z2     mean    SD                <<<---===  formatting pattern
     P(x <= 15) = normcdf(-9999, 15.5,  16,   3.57771) = 0.4444  (rounded).   



(d)  Do it as (b).



(e)  Do it as (a).

Solved.

---------------

If you are a beginner student in learning probability distributions, I advise you to start learning
this kind of computations using the online calculator, to which I referred above.

It provides a graphical support, so at each step you do understand what you are doing.
In addition, this graphical support prevents you from making mistakes.

When you learn enough this kind of computations, you can switch to your regular calculator,
but even then you may use the online calculator for checking purposes.



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