The average playing time of compact discs in a large collection is 37 min, and t

Question

The average playing time of compact discs in a large collection is 37 min, and the standard deviation is 6 min. (a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? 1 standard deviation above 43 1 standard deviation below 31 mean standard deviation above 49 he mean 2 standard deviation below 25 the mean (b) Without assuming anything about the distribution of times, at least what percentage of the times are between 25 and 49 min? (Round the answer to the nearest whole number) At least 75 (c) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than 19 min or greater than 55 min? (Round the answer to the nearest whole number.) No more than 8 x (d) Assuming that the distribution of times is normal, approximately what percentage of times are between 25 and 49 min? (Round the answers to two decimal places, if needed.) 95.45 Less than 19 min or greater than 55 min? 27 Less than 19 min? x 135

Explanation / Answer

Mean ( u ) =37
Standard Deviation ( sd )=6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
c.
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 19) = (19-37)/6
= -18/6= -3
= P ( Z <-3) From Standard Normal Table
= 0.0013
P(X > 55) = (55-37)/6
= 18/6 = 3
= P ( Z >3) From Standard Normal Table
= 0.0013
P( X < 19 OR X > 55) = 0.0013+0.0013 = 0.0027                  
0.27% are less than 19, greater than 55

d.
P(X < 19) = (19-37)/6
= -18/6= -3
= P ( Z <-3) From Standard Normal Table
= 0.0013 = 0.13%

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