Here is the main question:
Police estimate that 80% of drivers now wear their seatbelts. They set up a safety roadblock,
stopping cars to check for seatbelt use
If they stop 20 cars during the first hour, what’s the probability they find at least 18 drivers
wearing seat belts? Use a
normal
distribution to compute your answer
Here is the Answer: = 0.2005
My work:
I have been following this formula:
u = np (How many times a thing is done * success percent)
Y = npq (How many times a thing is done * success percent * (1 – success percent))
Then -> (what you want to measure) – (u) / sqrt(Y) = final Z-score
As you can see my answer is 0.83 and Ive been trying to figure out where I'm going wrong for a good 2 hours now..
Best Answer
I see you have set up the calculation $$ \frac{17.5 - 16}{\sqrt{3.2}}, $$ which looks like the correct approach to me.
Giving only two digits after the decimal place for a $Z$-score near $0.83$ is not enough precision to give a result accurate to four places, such as the probability $0.2005.$ You should carry more digits in your calculations.
Even if you did not increase the number of digits, you should be more careful how you round. The result should have been closer to $0.84$ than $0.83.$
Once you have a more accurate answer, you can look it up in a table of the normal distribution. Note that many such tables require you to add $0.5$ to the result in the table in order to get a probability, and in this case it will give you the probability of $17$ or fewer drivers wearing seatbelts.
I get $0.2009,$ so I suspect that the person writing the answer sheet reported more significant digits than their own methods justified.