please help and please type I can read some handwriting! CHAPTER 8 WORKsHEET CEN

Question

please help and please type I can read some handwriting!

CHAPTER 8 WORKsHEET CENTRAL LIMIT THEOREM (CLT) 1. The weight of potato chips in a medium-size bag is stated to 10 aunces. The amount that the packaging machines put in these bags is believed to follow a model with mean 10.2 ounces and standard deviation 0,12 ounces. (a) Identify individuals and variable here. Individuals variable (b) What proportion of all medium-size bags are underweight (ess than 10 ounces)? (e) Mike bought a potato chip bag of 10.6 ounce. What do you think he got unusually high? Explain your answer by the percentile of Mike's chip. (d) Some of the chipe are sold in "bargain packs" of 3 bags. What's the probability that the mean weight of 3 bags is below 10 aunces? (e) What's the probability that the mean weight of a 10 bag case of potato chips is below 10 ounces? (f What does the mean weight of 3 bags correspond to the 5th percentile?

Explanation / Answer

a. The individuals refer to each cases to be studied, that is potato chips and variable in question is weight, which can change values from case to case.

b. P(X<10)=P[Z<(10-10.2)/0.12] [use, Z=(X-mu)/sigma, where, X is raw score, mu is population mean, and sigma is population standard deviation]

=P(Z<-1.67)

=0.0475 (ans)

c. Substitute the raw score, X=10.6 in Z score formula to compute the Z score.

Z=(10.6-10.2)/0.12=3.33

The corresponding area is: 0.9995

Therefore, (1-0.9995)=0.0005 is proportion of scores below 10.6 ounce. The value is unusual (less than 5%).

d. P(X<10)=P[Z<(10-10.2)/(0.12/sqrt 3)] (according to CLT, with repeated samples of size N are drawn from any population with mean mu and standard deviation, sigma, then as N becomes large, the sampling distribution of sample means approach normality, with mean mu and standard deviation, sigma/sqrt N)

=P(Z<-2.89)=0.0019 (ans)

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