q19 ch4: solve step 5 please The average playing time of music albums in a large collection is 37 minutes, and the standard deviation is 6 minutes.(a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviationsaway from the mean?(b) Without assuming anything about the distribution of times, at least what percentage of the times are between 25 and 49 minutes?(c) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than19 minutes or greater than 55 minutes?(d) Assuming that the distribution of times is approximately normal, about what percentage of times are between 25 and 49 minutes?Less than 19 min or greater than 55 min? Less than 19 min?Step 1(a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviationsaway from the mean?To find the value that is some number of standard deviations above or below the mean, we will add or subtract some multiple of thestandard deviation, s, to the sample mean, x. The average playing time was given to be 37 minutes with a standard deviation of6 minutes, so we have x = 3737 and s = 6To find a value above the mean indicates we will use ladditionaddition. Finding a value below the mean indicates we will usesubtractionsubtraction. Finding values that are away from the mean includes values that are above and below the mean,indicating we will use addition and subtractionaddition and subtractionStep 2We know the mean is x = 37 and the standard deviation is s = 6. To find values above the mean indicates addition, values below the meanindicates subtraction, and both addition and subtraction will be used to find values away from the mean.Find the value that is 1 standard deviation above the mean.1 standard deviation above the mean = x + s-+5= 37 + 6= 4343Find the value that is 1 standard deviation below the mean.1 standard deviation below the mean = x - s= 37 - 6= 3131Find the values that are 2 standard deviations away from the mean.2 standard deviations above the mean = x + 25== 37 + 2 6649492 standard deviations below mean = x - 25= 37 - 2 6= 12525Step 3(b) Without assuming anything about the distribution of times, at least what percentage of the times are between 25 and49 minutes? (Round the answer to the nearest whole number.)When we assume nothing about the shape of the distribution, we will use Chebyshev's Rule to get a sense of the distribution of datavalues. Chebyshev's Rule will tell us the minimum percentage of observations that are within k standard deviations of the mean whenkz 1.100%1 -We previously determined that 25 is two standard deviations below the mean and 49 is two standard deviations above the mean. Thus, wewill use k- 22 in Chebyshev's Rule to determine the minimum percentage of times that are between 25 and 49 minutes.1100 1-% = 100 1-2= 7575 %Thus, at least 7575 % of times are between 25 and 49 minutes.Step 5We determined that 19 minutes and 55 minutes are 3 standard deviations from the mean. Thus, we will usek =inChebyshev's Rule to determine the minimum percentage of times that are between 19 and 55 minutes. (Round your final answer to thenearest whole number.)100 1-% = 10O 1-%However, we are asked to find the percentage of values that are less than 19 minutes and more than 55 minutes. These are values thatare found in the lower tail and in the upper tail of the distribution, not the middle. To find this number we will subtract the minimumpercentage of values that are between 3 standard deviations of the mean from 100%. (Round your answers to the nearest wholenumber.)100% -%=Thus, there will be no more than% of times less than 19 minutes or more than 55 minutes.Submit Skip (you cannot come back)Step 4(c) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either lessthan 19 minutes or greater than 55 minutes? (Round the answer to the nearest whole number.)Again, we are not assuming anything about the shape of the distribution, so we will continue using Chebyshev's Rule. We will determinethe minimum percentage of times that are between 19 minutes and 55 minutes and then determine the maximum percentage of timesthat are less than 19 minutes or greater than 55 minutes. First, we need to know how many standard deviations 19 minutes and55 minutes are from the mean so we can determine the value for k.We previously determined that 43 minutes and 31 minutes are 1 standard deviation from the mean and that 49 minutes and 25 minutesare 2 standard deviations from the mean.2531374349X- 3sX- 2sX-5X+5X+ 2sX+ 3sPerhaps 19 minutes and 55 minutes are 3 standard deviations from the mean. Let's investigate.3 standard deviations above the meanx + 3s = 37 + 3(6655553 standard deviations below the meanx - 3s = 37 - 3(6= 1919We showed that 19 minutes and 55 minutes areare 3 standard deviations from the mean.

We showed that 19 minutes and 55 minutes are 3 standard deviations from mean. Here, k=3

We determined that 19 minutes and 5s minutes are 3 standard deviations from the mean. Thus, we will use k = Chebyshev's Rule to determine the minimum percentage of times that are between 19 and 55 minutes. (Round your final answer to the nearest whole number.) in 100 1- = 100 1- However, we are asked to find the percentage of values that are less than 19 minutes and more than 55 minutes. These are values that are found in the lower tail and in the upper tail of the distribution, not the middle. To find this number we will subtract the minimum percentage of values that are between 3 standard deviations of the mean from 100%. (Round your answers to the nearest whole number.) 100% - ( ]% -( Thus, there will be no more than | % of times less than 19 minutes or more than 55 minutes. Submit Skip (you cannot come back)