Probabilty help needed?

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!

Probabilty help needed?
Without doing all of the calculations...





Calculate P(X %26lt; 20.857), which you already did in (A).





Then, use that probability in a binomial distribution modeling.
Reply:I'm not exactly sure myself. but i believe it is saying that out of y/100 of the mints weighs less then 20.857 grams.





Really hoped this helped!





(Sorry if it didn't)