Using

n,p

, as well as the values

a

and

b

from part (a), find the probability

P(a

, where

Y_(n)N(\\\\mu ;\\\\sigma ^(2))

, with mean

\\\\mu =n*p

, and variance

\\\\sigma ^(2)=n*p*(1-p)

(so that the standard deviation is

\\\\sigma =\\\\sqrt(np(1-p))

.\ Your probability

P(a

should be approximately equal to

P(a

(this is due to the Central Limit Theorem, which will be discussed soon). Write your answer (i.e. the value

P(a

) as

R

variable pclt , i.e.\ pclt

=

some expression>\ Do not round your answer.\ Hint: Use the R function pnorm (

q

, mean,

sd

, lower.tail = TRUE) . (see R documentation for details).

Using n,p, as well as the values a and b from part (a), find the probability P(aYnb), where YnN(;2), with mean =np, and variance 2=np(1p) (so that the standard deviation is =np(1p) ). Your probability P(aYnb) should be approximately equal to P(aSnb)=P(p^p Do not round your answer. Hint: Use the R function pnorm (q, mean, sd, lower.tail = TRUE). (see R documentation for details). Using n,p, as well as the values a and b from part (a), find the probability P(aYnb), where YnN(;2), with mean =np, and variance 2=np(1p) (so that the standard deviation is =np(1p) ). Your probability P(aYnb) should be approximately equal to P(aSnb)=P(p^p Do not round your answer. Hint: Use the R function pnorm (q, mean, sd, lower.tail = TRUE). (see R documentation for details)