I need help in my statistics class I don\'t know if am doing the right thing for

Question

I need help in my statistics class

I don't know if am doing the right thing for the second one.

Suppose we had a second bag of m&ms;, we would expect about 14% of the the second bag would be yellow. However, like the first bag,there is a chance that proportion will not equal O0.14 (or the proportion in the first bag,for that matter). Use the proportions from the first bag and from a new bag to determine what percent of bags of mam's size will have a yellow proportion between those two values Bag 1 Bag 2 Yellow Proportion % Bags Below Observed Sketch the two observed proportions on the normal distribution N(0.14,0.05) and note the percent of observations we would expect to see between the two observed proportions What About Peanut Butter m&m;'s? Do peanut butter m&m;'s follow the same color distribution as milk chocolate m&m;'s? If so, we would expect about 14% of the candies in a peanut butter m&m; bag would be yellow. would you be surprised if 15% were yellow? what about 20%2 30%? At what point would you suspect the color distribution for peanut butter m&m;'s may be differemt Open a bag of peanut butter måm's and note the proportion of yellow: Plot this value on the N(0.14,0.05) distribution and calculate the % of observations we'd expect to be more extreme than this observation. Based on this %, do you feel you have evidence to suggest the color distribution of peanut butter måm's may be different? Why or why not? Everything I Ever Wanted to Know about AP Statistics I Learned From a Bag of m&m;'s 11

Explanation / Answer

Yes, you are doing the right thing assuming that 0.05 is sd.The sample proportion of yellow candies is 11/57=0.1930 and the population proportion is normal with mean 0.14 and sd =0.05.So,z-score is=(0.1930-0.14)/0.05=1.06,with required probability = P(N(0,1)<1.06)=0.8554

But,in general N(mu,sigma^2) distribution means mean is mu and variance is sigma^2 and sd id sigma.Now,following the same notation,variance is 0.05,So,standard deviation is =sqrt{0.05}=0.2236.Then,the z-score would be z=(0.1930-0.14)/0.2236=0.23703,with required probability=P(N(0,1)<0.23703)=0.59368

I dont know the actual notation of the paper.Hence,I mentioned both.

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