Question

please help with b and c i dont know what im calculating wrong

An advertiser claims that the average percentage of brown MAM'S candies in a package of milk chocolate MONSI 14%. Suppose you randomly select a package of 51 candies and determine the proportion of brown candies in the package (a) What is the approximate distribution of the sample proportion of brown candies in a package that contains si candies? The distribution is binomial and cannot be approximated The distribution is approximated by a normal distribution with mean 7.1 and standard deviation 714 The distribution is approximated by a normal distribution with mean 0.14 and standard deviation 0.0486 The distribution is approximated by a normal distribution with mean 14 and standard deviation 0.0486. The distribution is approximated by a normal distribution with mean 0.14 and standard deviation 0.86. the S o n (b) What is the probability that the sample percentage of brown and 0.8485 x the d ance to four deca ) (c) What is the probability that the sample percentage exceeds 25 0.0119 x (Round your answer to forec ) (d) within what range would you expect the sample proportion to be about lower 0,04 upper limit 0.24 of them (Roundur answer to two deca )

Answer 1

Solution- compute 5 the probability that the sample of brown percentage is less than 1904 PLÊ <0.19) = Plze - A) Pu ) 0.17, PE 0.14 9=(1-) - 9-0.14) -0.86 n=51 PC ze op over one) -p (2210290) 51 TPCB 2019) = 0.84827 0.14086 c) P (P> 0 4) - - = 03 ] - plzz 0.95 -0.14 1 Jord: *0-56) 4P7777 #33720) =p(222.2639)= 0.017. [P(870.25)=0.017 d) 95% = 24=1.96. 950 Ра p + 1.96 10:14 YO.86 =) 0:14 +1.96 % 51 -0.044767, 0.23523 lower limit = 0.0447 upper limit = 0.2352.