Old Is Gold Question Solution(BBS 1st year)

 

  Q4.The standard deviation of a symmetrical distribution is 11. What must be the value of fourth moment about mean for the distribution to be normal? Solution:

 

 

 

 

 

  Q12. The difference between the upper quartile and lower   quartile of a certain frequency distribution is 4 and their sum is   16. Calculate the quartile deviation and its coefficient. Solution:

 

 

  Q8.The Mean and standard deviation of a series of 20 items are found to be 15 and 10  respectively. If an additional item of 5 included in the series, find the revised values of Mean  and standard deviation. Solution: Again,

 

 

 Q4.The coefficient of variation of a symmetrical distribution is   9% and mean of that distribution is 40. Find the value of variance. Solution:

  Q3.Coefficient of variation of a distribution is found to be 20% and variance of distribution is 49.what is the mean value of the distribution?   Solution:

  Q.2. Calculate the quartile deviation when the values of lower quartile and upper quartile are 12 and 18 respectively. Also, calculate the coefficient of quartile deviation. Solution:

 1.Coefficient of variation of a distribution is found to be 20% and variance of distribution  is 36.Find the mean value of the distribution . solution: Given, coefficient of variance(CV) = 20% variance = 36 Mean(X̅)= ? we know that, standard deviation( σ) =  √ variance                                          = √ 36                                          =6 CV =  σ/ X̅ *100 or, 20 =  σ /  X̅  *100 or, X̅ = 600/20 X̅=30 Hence Mean is 30.                     

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