Hi everyone. While I'm doing well with CLT problems based on means, I run into trouble when it comes to sums. My dilemma is simple: After I've converted to normal-distribution form, what do I do with my new standard deviation? Do I multiply it by N, as I do with the mean, or by the square root of N? Neither way seems to work consistently, which is seriously confusing me.
Here are two problems to illustrate what I mean.
-Problem A: "Letter envelopes are packaged in boxes of 100. It is known that on average the envelopes weigh 1 ounce, with a standard deviation of 0.05 ounces. What is the probability that 1 box of envelopes weighs more than 100.4 ounces?"
I got the right answer on this one, as follows: After converting to a normal distribution for a single envelope, I had E(X)=1 and standard deviation=0.005. Since the problems asks us to look at a whole box (100 envelopes) at a time, I solved as follows:
Pr(X>100.4) = 1 - Pr(X<100.4) = 1 - Pr(Z<([100.4-(1*100)]/[0.005*100]) = 1 - Pr (Z<0.4/0.5) = 1 - Pr (Z<0.8) =0.2119
As you can see, I just multiplied both the mean and the standard deviation by N=100 to get the correct solution. This approach worked perfectly for the next few problems, but completely failed for this problem:
-Problem B: "Let X and Y be the number of hours that a randomly selected person watches movies and sports, respectively, during a three-month period. The following is known about X and Y:
E(X)=50, E(Y)=20, Var(X)=50, Var(Y)=30, Cov(X,Y)=10
100 people are randomly selected and observed for these three months. If T is the total hours that they watch movies or sporting events during this time period, approximate the value of Pr(T<7100)."
Since Ti=Xi+Yi (where Ti/Xi/Yi are hours watched for a random person), I was able to find E(Ti)=70 and Var(Ti)=100, and thus s.d.=10. When I tried solving the same way as in Problem A, however, this is what happened:
Pr(T<7100)=Pr(Z<[7100-(70*100)]/[10*100])=Pr(Z<100/1000)=Pr(Z<0.1)=0.5398
In order to get the book solution of 0.8413 I would need Pr(Z<1), which would require me to multiply the standard deviation by the square root of N (i.e., 10) instead of just by N as I did in Problem A.
Why does the solution process change between these two problems? If I'm supposed to multiply standard deviation by sqrt(N), the first solution doesn't work; if I multiply by N instead, the second solution doesn't work. What am I missing?
Here are two problems to illustrate what I mean.
-Problem A: "Letter envelopes are packaged in boxes of 100. It is known that on average the envelopes weigh 1 ounce, with a standard deviation of 0.05 ounces. What is the probability that 1 box of envelopes weighs more than 100.4 ounces?"
I got the right answer on this one, as follows: After converting to a normal distribution for a single envelope, I had E(X)=1 and standard deviation=0.005. Since the problems asks us to look at a whole box (100 envelopes) at a time, I solved as follows:
Pr(X>100.4) = 1 - Pr(X<100.4) = 1 - Pr(Z<([100.4-(1*100)]/[0.005*100]) = 1 - Pr (Z<0.4/0.5) = 1 - Pr (Z<0.8) =0.2119
As you can see, I just multiplied both the mean and the standard deviation by N=100 to get the correct solution. This approach worked perfectly for the next few problems, but completely failed for this problem:
-Problem B: "Let X and Y be the number of hours that a randomly selected person watches movies and sports, respectively, during a three-month period. The following is known about X and Y:
E(X)=50, E(Y)=20, Var(X)=50, Var(Y)=30, Cov(X,Y)=10
100 people are randomly selected and observed for these three months. If T is the total hours that they watch movies or sporting events during this time period, approximate the value of Pr(T<7100)."
Since Ti=Xi+Yi (where Ti/Xi/Yi are hours watched for a random person), I was able to find E(Ti)=70 and Var(Ti)=100, and thus s.d.=10. When I tried solving the same way as in Problem A, however, this is what happened:
Pr(T<7100)=Pr(Z<[7100-(70*100)]/[10*100])=Pr(Z<100/1000)=Pr(Z<0.1)=0.5398
In order to get the book solution of 0.8413 I would need Pr(Z<1), which would require me to multiply the standard deviation by the square root of N (i.e., 10) instead of just by N as I did in Problem A.
Why does the solution process change between these two problems? If I'm supposed to multiply standard deviation by sqrt(N), the first solution doesn't work; if I multiply by N instead, the second solution doesn't work. What am I missing?
Central Limit Theorem
0 commentaires:
Enregistrer un commentaire