9. /9 points PeckDevStat7 4.E.038. My Notes Ask Your Teacher The average playing
ID: 3073525 • Letter: 9
Question
9. /9 points PeckDevStat7 4.E.038. My Notes Ask Your Teacher The average playing time of compact discs in a large collection is 36 min, and the standard deviation is 4 min. (a) What value is 1 standard deviation above the mean1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above the mean 1 standard deviation below the mean standard deviation above the mean 2 standard deviation below the mean (b) Without assuming anything about the distribution of times, at least what percentage of the times are between 28 and 44 min? (Round the answer to the nearest whole number.) At least (c) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than 24 min or greater than 48 min? (Round the answer to the nearest whole number.) No more than (d) Assuming that the distribution of times is normal, approximately what percentage of times are between 28 and 44 min? (Round the answers to two decimal places, if needed.) Less than 24 min or greater than 48 min? Less than 24 min? 9% Need Help? L.Readí.Ir Talk to a TutorExplanation / Answer
Solution: Given that mean = 36 min, standard deviation = 4
(a) u + 1s = 36 + 4 = 40
u - 1s = 36 - 4 = 32
u + 2s = 36 + 8 = 44
u - 2s = 36 - 8 = 28
(b)
28 = 36 - 2*4
44 = 36 + 2*4
k = 2
=> 1 - (1 / (k^2)) = 1 - 1/4 = 0.75 = 75 %
(c) Since k = 3
=> 1 - (1 / (k^2)) = 1 - 1/9 = 8/9 = 0.89
P(24<X<48) = 0.89
P(X<24) + P(X>48) = 1 - 0.89 = 0.11 = 11%
(d)
P(28<X<44)
= P((28-36)/4 <Z< (44-36)/4)
= P(-2<Z<2)
= 2P(Z<2) - 1
= 2*0.9772 - 1
= 0.9544
= 95.44 % (rounded)
P(X<24) + P(X>48)
= P(Z<-3) + P(Z>3)
= 2(1 - P(Z<3))
= 2(1 - 0.9987)
= 0.0026
= 0.26 % (rounded)
P(X<24) = P(Z<-3)
= 1 - P(Z<3)
= 1 - 0.9987
= 0.0013
= 0.13 % (rounded)
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