are you a chocolate "purist", or do you like other ingredients in your chocolate? american demographics reports that almost 75% of consumers like traditional ingredients such as nuts or caramel in their chocolate. they are less enthusiastic about the taste of mint or coffee that provide more distinctive flavors. a random sample of 200 consumers is selected and the number who likes nuts or caramel in their chocolate is recorded.

Question

(a) what is the approx. sampling distribution for the sample proportion p? what are the mean and standard deviation for this distribution?
(b) what is the probability that the sample percentage is greater than 80%?
(c) within what limits would you expect the sample proportion to lie about 95% of the time?

Answer 1

OUATTARA MAMANDOU
CCNY-CUNY
EXO: 7.47 page 281

P=0.75 Q=1-0.75 = 0.25 N=200
A- SE(^P)=sqrt (P.Q)/N where Q=1-P
SE(^P)=sqrt(0.75*0.25)/200
=0.306 SE(^P)=0.0306
Approximately normal with mean of 0.75 and standard deviation of 0.0306
B- Let find the Zscore of 80% z=^P-P/stddev
P(^P>80%) or P(^P>0.80)= P(0.80-0.75/0.0306)=1.63 ¡ú zvalue of 1.63 from the table is 0.9484 or P(z¡Ü1.63)=0.9484
Since P (^P>0.80)= 1-P(z¡Ü80). Then we compute 1-0.9484=0.0516.
So the probability that the sample proportion is greater than 80% is 0.0516
C- From the Empirical Rule and the general properties of the normal distribution, approximately 95% of the measurements will lie within 2 or 1.96 standard deviations of the mean:
P+-2SE since P=0.75 and SE(^P)=0.0306 we have 0.75+-2(0.0306)¡ú¡ú 0.75+-0.06
OR 0.75+0.06=0.81 and 0.75-0.06=0.69
Hence 95% lie between 0.69 and 0.81 OR 0.75+-0.06

Answer 2

Good job Mr Ouattara. Your answers are just excellent.
You are brillant.