Question 1205742
.
On some busy streets, diamond lanes are reserved for taxis, buses, and cars 
with three or more passengers. It is estimated that 20% of cars travelling 
in a certain diamond lane have fewer than three passengers. 
If 60 cars that travel in a certain diamond lane are selected at random, find the probability that:
exactly 10 cars have less than 3 passengers.
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<pre>
Obviously, this problem is on the binomial distribution.

The number of trials is n= 60; the probability of the success (fever than 3 passengers) is p= 0.2.
The number of success trials k is k=10.


Since the number of trials, 60, is great, it is recommended to use the normal approximation to 
the binomial approximation.

In this normal approximation, the mean is mean = np = 60*0.2 = 12; the standard deviation
SD = {{{sqrt(n*p*(1-p))}}} = {{{sqrt(60*0.2*(1-0.2))}}} = 3.098386677.


In this problem, you should find a normal distribution probability to get k=10. 
We can use a regular hand calculator TI-83/84 with standard function normalcfd

                   z1   z2   mean    SD          <<<---===  formatting pattern 
    P = normalcfd(9.5, 10.5, 12,   3.098386677).                                  (1)


Here 9.5, 10.5 are the continuity correction factors to the number of 10.


The  <U>ANSWER</U>  (the value of (1)) is the probability  P = 0.1043  (rounded).


Alternatively, to find the value of (1), you may use the online calculator at web-site

https://onlinestatbook.com/2/calculators/normal_dist.html


It has very convenient simple interface, so even a beginner student can use it without further
explanations.  This calculator will produce the same answer (the same value of probability).
</pre>

Solved.