- #1
bloynoys
- 25
- 0
Alright my brother posed this question to me tonight. You have 7 coins (for this sake we will go with the "normal" coins, penny, nickel, dime and quarter) and we are trying to find the probability that a person has over a dollar in coins.
So basically I did it and just wanted to confirm my answer before telling him tomorrow morning.
You just do:
E(x)=((1*(1/4))+(5*(1/4))+(10*(1/4))+(25*(1/4)))
E(x)= 41/5 = 10.25
7*E(x)=71.75
Var(x)=E(x^2)-E(x)^2
Var(x)=187.75-105.0625
Var(x)=82.6875
7*Var(x)=578.8125
Now I believe that I can do normal Z score stuff to find probability with the continuity correction. So we are looking for P(Z≥100) so we'll flip it into 1-P(Z≤100). Then we will add the continuity correction to make it 1-P(Z>100.5) and calculate it from there. Am I right so far?
So I go:
(71.75-100.5)/(24.058)
Z≈-1.195
And thus, using R normal probability calculator we are looking at a final answer of .1160379. Did I do that all right? Thanks!
So basically I did it and just wanted to confirm my answer before telling him tomorrow morning.
You just do:
E(x)=((1*(1/4))+(5*(1/4))+(10*(1/4))+(25*(1/4)))
E(x)= 41/5 = 10.25
7*E(x)=71.75
Var(x)=E(x^2)-E(x)^2
Var(x)=187.75-105.0625
Var(x)=82.6875
7*Var(x)=578.8125
Now I believe that I can do normal Z score stuff to find probability with the continuity correction. So we are looking for P(Z≥100) so we'll flip it into 1-P(Z≤100). Then we will add the continuity correction to make it 1-P(Z>100.5) and calculate it from there. Am I right so far?
So I go:
(71.75-100.5)/(24.058)
Z≈-1.195
And thus, using R normal probability calculator we are looking at a final answer of .1160379. Did I do that all right? Thanks!