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 43 the mean 1 standard deviation below 31 31 the mean 2 standard deviation above 49 49 the mean 2 standard deviation below 2 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 75 on withina

Explanation / Answer

(a) Mean = 37 min and Standard deviation = 6 min

so value 1 standard deviation above mean = Mean + standard deviation = 37 + 6 = 43

value 1 standard deviation below mean = Mean - standard deviation = 37 - 6 = 31

so value 2 standard deviation above mean = Mean + 2*standard deviation = 37 + 2* 6 = 49

value 2 standard deviation below mean = Mean -2 * standard deviation = 37 - 2* 6 = 25

(b) As we applied Chebyshev inequality :

The minimum values which are under k times standard deviation = (1-1/k2 )

so for k =2 or say values between 2 times standard deviation (25, 49) is = (1-1/4) = 3/4 = 75%

(c0 similarly if k =3 or say values are between 3 times standard deviation (19, 5) is = (1- 1/9) = 8/9

os the values will be no more than 1/9 times = 11%

(d) What percent of times are between 25 and 49 minutes = 95.45 (as per empirical rule and Z - table)

so Pr( Less than 19 min or greater than 55 min) = 1 - Pr(in between +-3 sigma value) = 1- 0.9970

= 0.30%

so Pr(Less than 19 min) = 0.30/2 = 0.15

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