The average playing time of cómpact discs in a large collection is 35 minutes, a
ID: 3063319 • Letter: T
Question
The average playing time of cómpact discs in a large collection is 35 minutes, and the standard deviation (a) What value is 1 standard deviation above th e mean? 1 standard deviation below the mean? What values are 2 standard deviations away frorn the mean? 1 standard deviation above the mean 1 standard deviation below the mean 2 standard deviations above the mean (b) Without assuming anything about the distribution of times, at least what percentage of the times are between 21 and 49 minutes? (Round the answer to the nearest whole number.) At least 0% (c) Without No more than t assurning anything about the distribution of times, what can be said about the percentage of times that are either less than 14 minutes or greater than 56 minutes? (Round the answer to the nearest wi d) Assuming that the distribution of times is normal, about what percentage of times are between 21 and 49 minutes? (Round the answers to two decimal places, if needed.) Less than 14 min or greater than 56 min? Less than 14 min?Explanation / Answer
Answer to the question is as follows:
Mean = 35
Stdev = 7
a.
1 std deviation above mean is 35+7 = 42
1 std deviation below mean is 35-7 = 28
2 std deviation above mean is 35+2*7 = 49
2 std deviation below mean is 35-2*7 = 21
b.so, 21 and 49 are 2 deviations away from mean. Chebyshev' rule states that 1-1/k^2 or 1-1/4 = 75% of area under curve/data points lie within 2 deviation of any normal curve
No more than 75%
c. P(X<14 or X>56) = P(Z<-3 or Z>3) = 1- P(-3<Z<3)
By empirical rule 3 deviation away from mean covers > = 1-1/3^2 = 88.9 % of the normal area
So, answer is : No more than 1-88.9% = 11.1%
d.
So, 21 and 49 are 2 deviations away from mean. Empirical rule states that 95% of area under curve/data points lie within 2 deviation of any normal curve
Empirical rules states that within deviation from mean area covered is 99.7%. So, Less than 14 min or greater than 56 min, is : 1 - 99.7% = .03%
Less than 14 min?
P(X<14) = P(Z<14-35 / 7) = P(Z<-3) = .015 or 1.5%
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