# 60 x σ σ n 10 100 010 p x 62 pz200 09772 c

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Chapter 8 / Exercise 22
Mathematical Excursions
Aufmann
Expert Verified
= 6.0 x σ = σ / n = 1.0/ 100 = 0.10 P( x <6.2) = P(z<2.00) = 0.9772 c. Probabilities concerning means do not apply to individuals. It is the information from part (a) that is relevant, since the helmets will be worn by one man at a time – and that indicates that the proportion of men with head breadth greater than 6.2 inches is 1 – 0.5793 = 0.4207 = 42.07%. 15. a. normal distribution μ = 69.0 σ = 2.8 P(x<72) = P(z<1.07) = 0.8577 b. normal distribution, since the original distribution is so x μ = μ = 69.0 x σ = σ / n = 2.8/ 100 = 0.28 P( x <72) = P(z<10.17) = 0.9999 Z 0 0.20 <--------------------------------| 0.5793 x 6.2 6.0 Z 0 2.00 <--------------------------------| 0.9772 x 6.2 6.0 _ Z 0 10.17 <--------------------------------| 0.9999 x 72 69.0 _ Z 0 1.07 <--------------------------------| 0.8577 x 72 69.0
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Chapter 8 / Exercise 22
Mathematical Excursions
Aufmann
Expert Verified
The Central Limit Theorem SECTION 6-5 169 c. The probability in part (a) is more relevant. Part (a) deals with individual passengers, and these are the persons whose safety and comfort need to be considered. Part (b) deals with group means – and it is possible for statistics that apply “on the average” to actually describe only a small portion of the population of interest. d. Women are generally smaller than men. Any design considerations that accommodate larger men will automatically accommodate larger women. 16. a. normal distribution μ = 0.8565 σ = 0.0518 P(x>0.8535) = P(z>-0.06) = 1 – 0.4761 = 0.5239 b. normal distribution, since the original distribution is so x μ = μ = 8565 x σ = σ / n = 0.0518/ 465 = 0.00240 P( x >0.8535) = P(z>-1.25) = 1 – 0.1056 = 0.8944 c. Whether the bags contain the labeled amount cannot be answered from the one bag examined. If each bag is filled with exactly 465 candies, then the company needs to make some adjustments – because then the probability a bag contains less than the claimed weight is 1 – 0.8944 = 0.1056 [and since 0.1056 > 0.05, it would not be unusual to get a bag with less than the claimed weight]. In practice, however, the bags are filled by volume or weight – and not by candy count [and several bags would have to be weighed to get an estimate of the bag-to-bag variability]. 17. a. normal distribution μ = 143 σ = 29 P(140<x<211) = P(-0.10<z<2.34) = 0.9904 – 0.4602 = 0.5302 -0.06 Z 0 <----------- 0.4761 x 0.8565 0.8535 -1.25 Z 0 <----------- 0.1056 x 0.8565 0.8535 _ Z 0 2.34 -0.10 <--------------------------------| <----------- 0.4602 0.9904 x 140 143 211
170 CHAPTER 6 Normal Probability Distributions b. normal distribution, since the original distribution is so x μ = μ = 143 x σ = σ / n = 29/ 36 = 4.833 P(140< x <211) = P(-0.62<z<14.07) = 0.9999 – 0.2676 = 0.7323 c. The information from part (a) is more relevant, since the seats will be occupied by one woman at a time. 18. a. normal distribution μ = 5.670 σ = 0.062 P(5.550<x<5.790) = P(-1.94<z<1.94) = 0.9738 – 0.0262 = 0.9476 If 0.9476 of the quarters are accepted, then 1 – 0.9476 = 0.0524 of the quarters are rejected. For 280 quarters, we expect (0.0524)(280) = 14.7 of them to be rejected. b. normal distribution, since the original distribution is so x μ = μ = 5.670 x σ = σ / n = 0.062/ 280 = 0.00371 P(5.550< x <5.790) = P(-32.39<z<32.39) = 0.9999 – 0.0001 = 0.9998