μ = np = (100)(0.5) = 50
σ = √(npq) = √(100*0.5*0.5) = √25 = 5

Use TI-84

normalcdf(45,55,50,5)

Answer: 0.6826894809

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

Let Y be the number of ones which occur when a die is rolled 300 times
Find P(45 < X < 55) 
μ = np = (300)(1/6) = 50
σ = √(npq) = √(300*(1/6)*(5/6)) = √41.66666667 = 6.454972244

Use TI-84

normalcdf(45,55,50,6.454972244)

Answer: 0.5614221135

Edwin


 

 

Answer by ikleyn(53765)   (Show Source): You can put this solution on YOUR website!
 .

Let x be the number of heads which occur when a coin is tossed 100 times 

and Y be the number of ones which occur when a die is rolled 300 times.

Find P(45 < X < 55) and P(45 < Y < 55)

~~~~~~~~~~~~~~~~







Both are typical binomial distributions.







Since the number of trials is comparatively large,  it is recommended to use 

the normal approximation to binomial distribution.







Go to web-site  https://mathcracker.com/normal-approximation-binomial-distribution



and find there online calculator specialized for such calculations.







It has perfect description,  clear instructions and simple and convenient input.



The output includes full theory/explanations with detailed step-by-step calculations shown,

and with every step commented.



It also provides graphical illustration of normal curve.







The calculator gives    P(45 < X < 55) = 0.6319        (with p = 0.5)





                                                      P(45 < Y < 55) = 0.5143        (with p = 1/6 = 0.16666)







In parallel with computing,  learn the relevant theory from this web-site.







//////////







Dear Edwin,





you can see some difference between my answer and your calculations.





I don't know precisely, what is the source of this difference.





I only know that they in the site refer to some "continuity correction factor",

which they do apply, while you do not.





I am not a specialist in it.



But I think that it would be interesting to learn in some way which methodology is correct.





Yesterday, very similar problem came to the forum and was solved by @Boreal under this link





https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1189215.html





@Boreal also applied this "continuity correction factor".





I performed similar calc with the solver for that condition, and obtained the same number/(answer) as @Boreal, with 4 decimal figures.





Please do not consider my post as an attempt to disprove or challenge your solution.





It is only my request to specialists to find the truth.







................ (written 1.5 hours later)







Edwin,  I made several numerical experiments,  and it seems to me that I found the source of discrepancy.  





normalcdf(45.5,54.5,50,5) = 0.6319, precisely as with the online calculator,



                                                        where I assigned the lower and upper limits as 46 and 54.





normalcdf(45.5,54.5,50,6.454972244) = 0.5143, precisely as with the online calculator,



                                                        where I assigned the lower and upper limits as 46 and 54.







So,  it looks like my question is resolved,  and the online calculator is correct.





The matter is in assigning upper and lower limits in the function  normalcdf  of pocket  (physical)  calculators.







 

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