0
$\begingroup$

The average weight of a set X of 100 bags of rice is 90 pounds , and the standard deviation is 8 pounds. Bag A weighs 2 standard deviations below the average weight and Bag B weighs 5 pounds more than Average.

Choose if quantity A or quantity B is bigger , equal or the information provided is not sufficient:

Quantity A: Difference between weight of Bag A and Bag B Quantity B: The range of weight of bags in set X

$\endgroup$
  • $\begingroup$ using formula : mean+3*(sd) - mean - 3*(sd) where sd = standard deviation .. i used this formula but the answer said range cannot be calculated.. $\endgroup$ – Dinesh Pabbi Jan 3 '18 at 11:58
0
$\begingroup$

Hint:

Without knowing how the data are distributed, how can we determine the range (to calculate quantity B) just from knowing the mean and standard deviation?

So, I feel the answer should be: Cannot be determined.

$\endgroup$
  • 1
    $\begingroup$ We cannot calculate quantity B exactly, but can we establish bounds on it? Clearly $B>0$, otherwise the standard deviation would be $0$. Also $B<90\cdot 100$ (99 bags of weight 0, one bag of weight $90\cdot 100$). Using all information in the question (weights are positive, mean=90, sd=8, one bag weighs 90-2 sd, another weighs 90+5), how tight could we make the bounds on B? Tight enough to decide if A is bigger? $\endgroup$ – Wouter Jan 3 '18 at 11:41
  • $\begingroup$ Hey your answer is right but cant we determine the range by using the standard deviation graph? And then using formula : mean+3*(sd) - mean - 3*(sd) where sd = standard deviation? i thought like this but your answer is right i still don't get why we can't calculate range? $\endgroup$ – Dinesh Pabbi Jan 3 '18 at 11:53
  • 1
    $\begingroup$ The formula you give, mean+3*(sd) - (mean - 3*(sd)) = 6*sd , is not a formula that gives you the range. Consider $\{0,2\}$: mean=1, sd=$\sqrt{2}$, range=2, 6sd$\neq$range. Or consider the beta distribution, which has range $[0,1]$ no matter the value of its standard deviation. $\endgroup$ – Wouter Jan 3 '18 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.