Can you explain the answers for part b? I don\'t quite understand how he arrives

Question

Can you explain the answers for part b? I don't quite understand how he arrives at the answer. This question is solved using chebychev's inequality.

The weight of a bag of a certain brand of potato chips is given as 6 ounces, but in fact it is a random variable whose expected value is 6.2 ounces and whose standard deviation is 0.17 ounce. Nothing else about the distribution is known. (a) What can you say about the probability that the weight of a randomly selected bag is between 6 ounces and 6.4 ounces? P(6 X 6.4) 1 0.172/0.22. (b) Find an interval in which you can be sure that 90% of the bag's weights will be found. (5.663 6.737)

Explanation / Answer

b)here fromCheby chev;s (1-1/k2 ) =0.90

where K= margin of error E/std deviation

hence 1/k2 =1-0.9=0.1

k2 =1/0.1=10

margin of error =std deviation*(10)1/2 =0.17*(10)1/2 =0.537

hence 90% confidence interval =mean -/+ margin of error =6.2-/+0.537 = 5.663 ; 6.737

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