Hi, I\'m really having a hard time with these 2questions. I\'ve been struggling
ID: 2953949 • Letter: H
Question
Hi, I'm really having a hard time with these 2questions. I've been struggling with them for so long that I amstarting to fall behind. Thanks for your help! 1.Assume thatthe population of heights of male college students is approximatelynormally distributed with mean m of 68 inchesand standard deviation sof 3.75 inches. A random sample of 16heights is obtained. Show all work.(A) Find the proportion of malecollege students whose height is greater than 70 inches.
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(B) Find the mean and standarderror of the xdistribution
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(C) Find P(x<70) 2. Two jelly beans areselected, one at a time from a bowl containing 10 black, 10 red and10 green jelly beans. Let x represent the number of black jellybeans selected in 2 draws from the bowl.
(A) If this experiment iscompleted without replacing the jelly beans, explain why x is not abinomial random variable.
(B) If this experiment iscompleted with replacement of the jelly beans, explain why x is abinomial random variable.
Hi, I'm really having a hard time with these 2questions. I've been struggling with them for so long that I amstarting to fall behind. Thanks for your help! 1.Assume thatthe population of heights of male college students is approximatelynormally distributed with mean m of 68 inchesand standard deviation sof 3.75 inches. A random sample of 16heights is obtained. Show all work.
(A) Find the proportion of malecollege students whose height is greater than 70 inches.
_
(B) Find the mean and standarderror of the xdistribution
_
(C) Find P(x<70) 2. Two jelly beans areselected, one at a time from a bowl containing 10 black, 10 red and10 green jelly beans. Let x represent the number of black jellybeans selected in 2 draws from the bowl.
(A) If this experiment iscompleted without replacing the jelly beans, explain why x is not abinomial random variable.
(B) If this experiment iscompleted with replacement of the jelly beans, explain why x is abinomial random variable.
Explanation / Answer
(1) Given X~Normal(=68, =3.75), n=16 (a) P(X>70) = P((X-)/ > (70-68)/3.75) =P(Z>0.53) =0.7019 (check normal table) (b) mean standard error is /n = 3.75/sqrt(16) = 0.9375 (c) P(xbar < 70) =P((xbar-)/(/n)Related Questions
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