The average playing time of compact discs in a large collection is 37 minutes, a

Question

The average playing time of compact discs in a large collection is 37 minutes, and the standard deviation is 6 minutes. 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 Without assuming anything about the distribution of times, at least what percentage of the times are between 25 and 49 minutes? (Round the answer to the nearest whole number.) At least 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 55 minutes? (Round the answer to the nearest whole number.) No more than Assuming that the distribution of times is normal, about what percentage of times are between 25 and 49 minutes? (Round the answers to two decimal places, if needed.) Less than 19 min or greater than 55 mm? Less than 19 min?

Explanation / Answer

b.
Chebyshev’s theorem                  
Step 1 : Subtract the Mean from the larger value.
= 49 - 37
Step 2 : Divide the diffrence by the standard deviation to get k
k = 12 / 6 = 2
Step 3 : Use Chebyshev’s theorem to find the percentage.
1 - 1/ k^2 = 1 - 1/4 = 1 - 0.25 = 0.75 OR 75 %   

c.
Step 1 : Subtract the Mean from the larger value.
= 55 - 37
Step 2 : Divide the diffrence by the standard deviation to get k
k = 18 / 6 = 3
Step 3 : Use Chebyshev’s theorem to find the percentage.
1 - 1/ k^2 = 1 - 1/9 = 1 - 0.1111 = 0.8889 OR 88.89 %                   

P( X < 19 OR X > 55) = 1 - 0.8889 = 0.1111 ~ 11.11 ~ 12%

d.

Mean ( u ) =37
Standard Deviation ( sd )=6
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 25) = (25-37)/6
= -12/6 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 49) = (49-37)/6
= 12/6 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(25 < X < 49) = 0.97725-0.02275 = 0.9545 ~ 96%          

e.
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                  

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