1. MARS, the company that makes m&m’s, claims that 20% of the candies it produce

Question

1. MARS, the company that makes m&m’s, claims that 20% of the candies it produces are orange.

(1)If you took a simple random sample of 75 m&m’s, what is the probability that your sample would have between 17% and 25% orange m&m’s?

Please round your final answer to FOUR decimal places.

(2)Find the quartiles of the sampling distribution of the sample proportion of orange m&m’s for samples of size 75.

2. In 2012, incoming freshmen had a mean high school GPA of 3.63. Let’s assume a standard deviation of 0.31. We are going to randomly sample 50 freshmen.

Determine the proportion of all samples of 50 freshmen that have GPAs within 0.05 points of the population mean GPA of 3.63. Round your answer to four decimal places.

Explanation / Answer

Approach: We approximate the distribution to normal distributions and calculate the probabilites

1. Probability of Orange candy, p = 20% = 0.2

Probability candy is not orange, q = 1-p = 0.8

This is a binomial distribution with p = 0.2 and q = 0.8

To approximate this to a normal distribution for a sample size n =75. np and nq should be at least 5

np = 75*0.2 = 15

nq = 75*0.8 = 60 both satisfy the condition hence this sample can be approximated to follow a normal distribution

mean , mu = np = 15

standard deviation = square root(npq) = sqrt(75*0.2*0.8) = sqrt(12) = 3.4641

Probability that your sample would have between 17% and 25% orange m&m’s

17% of 75 = 12.75

25% of 75 = 18.75

z score of 12.75 = (12.75-mu)/sd = (12.75-15)/3.4641 = -0.6495

z score of 18.75 = (18.75-15)/3.4641 = 1.0825

Hence Probability of the sample would have 17% to 25 % orange candies = 1-P(z<-0.6495)- P(z>1.0825)

P(z<-0.6495) from normal distribution table = 0.2580

P(z>1.0825) = 1- P(z<1.0825) = 1 - 0.8605 = 0.1395

Hence, Probability of the sample would have 17% to 25 % orange candies = 1-0.2580 - 0.1395 = 0.6025

(2)

Quartiles are values at probabilities 0.25. 0.50 and 0.75

Lets get the z scores from normal distribution table

1st quartile z score for (P= 0.25) = -0.6745

2nd quartile z score for (P=0.5)= 0 ( nothing but mean)

3rd quartile z score for (P=0.75) = 0.6745

lets calculate the no of ornage candies for each quartile from z score :

1st quartile = z*sd + mean = -0.6745*3.4641 + 15 = 12.6635 ~ 13 candies

2nd quartile is mean which is 15 candies

3rd quartile = 0.6745*3.4641 + 15 = 17.336 ~ 17 candies

Qn 2. mean = 3.63 and standard deviation = 0.31

As sample size is 50 , we can approximate this to a normal distribution N(3.63,0.31)

Z score of 0.05 points away from mean =0.05/0.31 = +/- 0.1613

P(score within 0.05 points of mean) =1- P(z<-0.1613) - P(z>0.1613)

To calculate the probability we are substracting the probabilities of all other outcomes from 1

= 1 - 0.4359 - 0.4359 = 0.1282

P(score within 0.05 points of mean) = 0.1282

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