Average of x^2 is= 1.67"Calculating Mean & STDev of # of Quarters

[noreturn2]

Homework Statement

Mind mean and STDev of number of quarters someone has on them at one time
0 - 87%
1 - 8%
2 - 3%
3- 1%
4 - 1%

The first time I did this I assumed a population but she said that was wrong and not what she was looking for.

Here was my new attempt:
x^2f(x)=
0(.87)
1(.08)
4(.03)
9(.01)
16(.01)

sum= .45

so: .45-(.2)^2=.6403

I want to do E[x^2]-[E[x]]^2=stdev

 

[FactChecker]

I guess I am missing something. Where are those multipliers (0, 1, 4, 9, 16) coming from? What formula are you using?

Oh, I guess I see. You have a variable with values (0, 1, 2, 3, 4) with probabilities 0.87, 0.08, 0.03, 0.01, 0.01, respectively. You should have made that clear in your problem statement.

In that case, you should be more careful about your calculations. I get 0.45-(0.21)^2 = 0.4059

 

[noreturn2]

I guess I am missing something. Where are those multipliers (0, 1, 4, 9, 16) coming from? What formula are you using?

I forgot to add in part of the problem statement. I've added that.

x^2f(x)

 

[vela]

so: .45-(.2)^2=.6403

$$0.45-0.2^2 = 0.45-0.04 = 0.41$$ So don't write that it's equal to 0.6403. The equal sign doesn't mean "the next step in my calculation gives me..." Trust me, when you write something like that, it annoys your teacher at best and probably drives her crazy. Instead, you could say $0.45-0.2^2 = 0.45-0.04 = 0.41$ for the first step and then $\sigma = \sqrt{0.41}=0.6043$.

(As @FactChecker noted, you should recheck your calculations.)

I want to do E[x^2]-[E[x]]^2=stdev

Again, this isn't true. The lefthand side is equal to the variance, not the standard deviation.

It appears you basically know what you're doing, but you need to be more careful in writing stuff down to demonstrate that to others (like the person who is going to grade your test).

 

[noreturn2]

$$0.45-0.2^2 = 0.45-0.04 = 0.41$$ So don't write that it's equal to 0.6403. The equal sign doesn't mean "the next step in my calculation gives me..." Trust me, when you write something like that, it annoys your teacher at best and probably drives her crazy. Instead, you could say $0.45-0.2^2 = 0.45-0.04 = 0.41$ for the first step and then $\sigma = \sqrt{0.41}=0.6043$.

(As @FactChecker noted, you should recheck your calculations.)


Again, this isn't true. The lefthand side is equal to the variance, not the standard deviation.

It appears you basically know what you're doing, but you need to be more careful in writing stuff down to demonstrate that to others (like the person who is going to grade your test).

So my answer of .64043 was correct?

 

[vela]

Close, but as @FactChecker noted, you need to check your calculations.

 

[noreturn2]

Close, but as @FactChecker noted, you need to check your calculations.

So what is throwing me off is I think my calculation of the mean is incorrect. Because more people has 0 quarters then say 2 quarters, the the mean should theoretically be less then 2 correct?

 

[vela]

Right. Since the vast majority of people had no quarters, you'd expect the mean to be relatively close to 0.

 

[noreturn2]

Right. Since the vast majority of people had no quarters, you'd expect the mean to be relatively close to 0.

So would it be .45/100%= .0045, which seems also wrong but makes sense. Since the .45 is the number of quarters in the "system" sort of say, and then we accounted for 100% of the population.

 

[FactChecker]

So what is throwing me off is I think my calculation of the mean is incorrect. Because more people has 0 quarters then say 2 quarters, the the mean should theoretically be less then 2 correct?

Yes. But your calculated mean of 0.2 (should be more accurate 0.21) is much smaller than 2. Why do you doubt that answer for the mean?

 

[FactChecker]

So would it be .45/100%= .0045, which seems also wrong but makes sense. Since the .45 is the number of quarters in the "system" sort of say, and then we accounted for 100% of the population.

I don't understand where you are getting the 0.45 number. In any case, dividing by 100% makes no sense to me.

I suggest that you stick with the established formulas and be more careful with them. You essentially got the right answer the first time, but you presented it carelessly and it was not possible (for me) to follow what you were doing.

 

[noreturn2]

Ok so here is what I have let me know if this looks better

Mean: .21
87(0)+8(1)+3(2)+1(4)+1(3)=21/100

STDev:
E[x^2]-[E[x]]^2

.45-(.21)^2=.4059
sqrt(.4059)

STDEVA: .637

 

[FactChecker]

Much better. Some improvements could be: Show the calculation of E[x^2] that gave you .45, label the variance, and use the same name STDev (not STDEVA).

 

[Ray Vickson]

Ok so here is what I have let me know if this looks better

Mean: .21
87(0)+8(1)+3(2)+1(4)+1(3)=21/100

STDev:
E[x^2]-[E[x]]^2

.45-(.21)^2=.4059
sqrt(.4059)

STDEVA: .637

Never write 87(0)+8(1)+3(2)+1(4)+1(3)=21/100; in fact, 87(0)+8(1)+3(2)+1(4)+1(3)=21, period. I think what you want to do is note that 0.87(0)+0.08(1) + 0.03(2) + 0.01(4) + 001(3) = 0.21, so that is what you should write. Alternatively, you can say in words that 87(0)+8(1)+3(2)+1(4)+1(3)=21 is 100 times the mean, so you need to divide the result by 100 to get the mean.