SOLUTION: A sample of n=6 BMI scores of a group of individuals has a mean of M = 24.5. One new BMI score is added to the sample and the new mean is found to be M = 25.0. What can you conclud

Algebra ->  Probability-and-statistics -> SOLUTION: A sample of n=6 BMI scores of a group of individuals has a mean of M = 24.5. One new BMI score is added to the sample and the new mean is found to be M = 25.0. What can you conclud      Log On


   



Question 1186994: A sample of n=6 BMI scores of a group of individuals has a mean of M = 24.5. One new BMI score is added to the sample and the new mean is found to be M = 25.0. What can you conclude about the value of the new BMI score?
Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
.
A sample of n=6 BMI scores of a group of individuals has a mean of M = 24.5.
One new BMI score is added to the sample and the new mean is found to be M = 25.0.
What can you conclude about the value of the new BMI score?
~~~~~~~~~~~~~~

Let  a,  b,  c,  d,  e,  and  f  represent the scores of the 6 participants.


You are given that

    %28a+%2B+b+%2B+c+%2B+d+%2B+e+%2B+f%29%2F6 = 24.5.


It means that  a + b + c + d + e + f = 6*24.5 = 147.      (1)



Next, they ask you to find the unknown score x from this equation

    %28a+%2B+b+%2B+c+%2B+d+%2B+e+%2B+f+%2B+x%29%2F7 = 25.0


From the last equation

    a + b + c + d + e + f + x = 7*25.0 = 175.             (2)


Replace here the sum  (a + b + c + d + e + f)  by 147, based on (1).  You will get

   147 + x = 175.


Hence,

         x = 175 - 147 = 28.


ANSWER.  The new BMI score is 28.

Solved.

--------------------

It is a typical problem on average scores.

To see many other similar (and different) problems solved, look into the lessons
    - Solved problems on average scores, weight, height and temperature
    - Solved problems on average scores
    - Solved problems on average age
    - Miscellaneous problems on average values
in this site.

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher.