Question

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 50 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.2 pounds, what is the probability that the sample mean will be each of the following? Appendix A Statistical Table a. More than 60 pounds b. More than 57 pounds c. Between 56 and 58 pounds d. Less than 53 pounds e. Less than 47 pounds (Round the values of to 2 decimal places. Round your answers to decimal places Type here to search oe ? 4 1:06 AM 1717/2019

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Answer 1

Soli Given that, The average of Aluminium, American uses (M) = 56.8 pounds. Standard deviation of annual usage (~) = 12.2 pounds. Random sample of American house holds (n) = 50. The distribution of sample mean I is normal distribution, mean llx = l = 56.8 Here the problem is about the calculation of probabilities by using the area property of Normal distribution. Here we use the standard normal variate z to find the probability of given event. mean ur Ulx = 56.8 standard deviation, oq = -2 3) VE = hten

> = 1.7253 :. The a-score for ř is, 2 = e-lz = -> 2= - 56.8 -0 1.7253 (a) More than 60 pounds. The poobability (& more than 60 pounds), > p (7>60) = P(2>185) = 0.03216 0.0322 Since, Z-Score - 60-56.8 - 1.7253 → Z-Score = 1.85 Cumulative asta auf 1$$

6) More than 57 pounds. The probability (I more than 57 pounds), P(Z >57) 2- Score = 57-56.8 1.7253 = 2-score = 0.19 O .14 >> P(Z >57) = p (230.12) a P[7357) =0.4599 (6) Between 56 and 58 pounds. The probability (I beteveen 56 and 58 pounds), - P( 36 5 7 258) If x=56 2 cose 56 - 56.8 77253 > Z - Score = -0.46

If 2 = 58 58-56.8 then 2-score = . 1.7253 -2 > 2-score = 0.69 SCOYC = 0.69. MA -0.46 0.69 Now, - P( 36 = 2 258) = P(-0.46 = 2 20.69) = P(250169) – P(25-0.46) 20. 7549 -0.32.28 p(365ăs58) =0.4324 (d) Less than 53 pounds. The probability (2 less than 53 pounds), P(53) 2-Score - 53-56.8 - 1.7253 > Z-Score - 2.20 >> PC Z 253) = P(22-220)

>> P(Z <53) = 0.0139] (e) Less than 47 pounds The probability (a less than a P(x247) 47 pounds), 2-Score = 47-56.8 1.7253 5.6% 0 2. Score = -5.68 => P(2247) = P(22-5.68) => p (2247) = 0)