Can anyone advise me on how to solve this statistics problem?

Suppose only 40% of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that between 180 and 230 (inclusive) of the drivers in the sample regularly wear a seat belt?





If you can help, it would be appreciated. I am not just looking for the answer, I want to understand how to do it too. Thanks!Can anyone advise me on how to solve this statistics problem?
Mean = (.4)(500) = 200





Standard Deviation = Sqrt[500(.4)(.6)] = 10.95





Normalizing the data given yields:





(180 - 200) / 10.95 %26lt;= z %26lt;= (230 - 200) /10.95





-1.8265 %26lt;= z %26lt;= 2.7397





Probability = .9630Can anyone advise me on how to solve this statistics problem?
Mean = 500(0.40) = 200


Variance = 500(0.4)(0.6) = 120


Standard deviation = sqrt(120)=10.9544


渭 = 200


蟽 = 10.9544


P( 180 %26lt; x %26lt; 230) =


P[( 180 - 200) / 10.9544 %26lt; X %26lt; ( 230 - 200) / 10.9544]


P( -1.8258 %26lt; Z %26lt; 2.7386) = 0.4664+0.4969 =0.9633


(Using the Normal probability table)