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The average playing time of compact discs in a large collection is 34 minutes, a

ID: 3069042 • Letter: T

Question

The average playing time of compact discs in a large collection is 34 minutes, and the standard deviation is 5 minutes. (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 the mean 1 standard deviation below the mean 2 standard deviations above the mean 2 standard deviations below the mean (b) Without assuming anything about the distribution of times, at least what percentage of the times are between 24 and 44 minutes? (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 19 minutes or greater than 49 minutes? (Round the answer to the nearest whole number.) No more than (d) Assuming that the distribution of times is normal, about what percentage of times are between 24 and 44 minutes? Round the answers to two decimal places, if needed.) Less than 19 min or greater than 49 min? Less than 19 min?

Explanation / Answer

SolutionA

mean=34

sd=5

1 std deviation above mean=mean+sd=34+5=39

1 std deviation below mean=mean-sd=34-5=29

2 std deviation above mean=mean+2sd=34+2*5=34+10=44

2 std deviation below mean=mean-2sd=34-2*5=34-10=24

Solutionb:

mean-2sd and mean+2sd=24 and 44

about 95% of the values fall within two standard deviations of the mean

ANSWER:95%

Solutinc:

mean-3sd=34-3*5=19

and mean+3sd=34+3*5=34+15=49

About 99.7% of all values within 3 standard deviations of the mean

ANSWER:99.7%

Solutiond:

use rcode:

pnorm(44,mean=34,sd=5)-pnorm(24,mean=34,sd=5)

0.9544997*100

=95.45%

Less tha 19 min or greater than 49 min'

pnorm(19,mean=34,sd=5)+pnorm(49,mean=34,sd=5,lower.tail=FALSE)

=0.002699796*100

=0.27%

less than 19 min

pnorm(19,mean=34,sd=5)

0.001349898*100

=0.13%

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