Question 1208601
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Answer: <font color=red>0.17879</font>


Explanation


We have a binomial distribution with these parameters
n = 525 = sample size
p = 0.06 = probability of success


To ensure we can use a normal distribution to approximate this binomial distribution, we must check that n*p > 5 and n*(1-p) > 5
n*p = 525*0.06 = 31.5
n*(1-p) = 525*(1-0.06) = 493.5
Both results exceed 5, so the requirements are met.
Some textbooks will raise the threshold to n*p > 10 and n*(1-p) > 10. The results also meet these requirements.


mu = n*p = 525*0.06 = 31.5 is the mean
sigma = sqrt(n*p*(1-p)) = sqrt(525*0.06*(1-0.06)) = 5.44150714 is the approximate standard deviation


Using a <a href="https://www.statisticshowto.com/what-is-the-continuity-correction-factor/">continuity correction factor</a>, we'll adjust P(X < 27) to get P(X < 26.5)


Compute the z score.
z = (x - mu)/sigma
z = (26.5 - 31.5)/5.44150714
z = -0.91886307807
z = -0.92 is the approximate z score when rounding to 2 decimal places.


Then use the <a href="https://www.ztable.net/">Standard Normal Distribution Table</a> (also known as Z table) that your teacher has provided you.
Usually such a table is found in the back of your stats textbook.


Locate the row that starts with -0.9
Highlight the column that has 0.02 at the top.
The value <font color=red>0.17879</font> is in this row and column. 
This value is approximate.
We can say that P(Z < -0.92) = <font color=red>0.17879</font> which is the approximate final answer in decimal form.
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