Probability answer verification and help?

i have solutions for parts A and C, im hoping you can help me with part B





the question is:





mints have a label weight of 20.4 grams. assume the distribution is N(21.37,0.16)





A) let X denote the weight of a single mint seleted at random, find P(X%26lt;20.857)





i got P(20.857-21.37/0.4) = 0.1003 from normal table





C) let Xbar equal the sample mean of 100 mints selected at random and weighed. find P(21.31%26lt;=Xbar%26lt;=21.39





my answer:


with n=100 the distribution becomes (21.37, 0.16/100)





P(21.31-31.37/ sqrt(.0016)) %26lt;= 21.39-21.37/sqrt(.0016)


= 0.6247 from normal tables. does that seem right?





this is the one i dont understand, part B





B)


100 mints are selceted at random and weighed. let Y equal the number of these mints that weigh less that 20.857 grams. approximate (Y%26lt;=5)





thank you!

Probability answer verification and help?
A is correct, good job.


C you get the proper value, good job. Your writing is a little obscure though. P(21.31%26lt;=Xbar%26lt;=21.39) = Φ((21.39-21.37)/√(0.16/100)) - Φ((21.31-21.37)/√(0.16/100)).


62.47% is correct.





B. You know the probability a mint weighs under 20.857 g is 10% (cf. A).


The probability that 5 or fewer mints in a sample of size 100 weigh under 20.857g is then:


P(Y%26lt;=5) = P(Y=0)+ P(Y=1)+ P(Y=2)+ P(Y=3) +P(Y=4) +P(Y=5)


This is a binomial distribution. P(Y=n) = p^n*(1-p)^(100-n)*100Cn


where p=0.1 (the probability a mint weighs less than 20.857g) and 100Cn = 100!/(n!(100-n)!


The correct result (computed using a VBscript program) yields P(Y%26lt;=5)= 5.7%


You can approximate that result using the normal approximation for the binomial distribution.


The equivalent normal distribution is N(np, (√(np(1-p))²) where n is the size of the sample and p your probability (10%).


It is considered valid when np%26gt;=10. here, np = 10, so we should be ok.


The apporximation is done using a normal distribution of N(10, 3²)


Also, when you consider your upper bound (here you have to find P(Y%26lt;=5)), you do a continuity correction because you are switching from a discrete distribution (binomial) to a continuous distribution (normal). You add 0.5 to your limit.


The approximation becomes: P(y%26lt;=5) = Φ((5.5-10)/3)) = Φ(-1.5) = 6.7% Not an ideal approximation, but not bad either.