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Each student spends money on snacks after school every afternoon:

Student 1's spending follows a normal distribution, he spends \$8 on average with a St. Dev of $3

Student 2's spending follows a normal distribution, he spends \$9 on average with a St. Dev of $2

Student 3's spending follows a normal distribution, he spends \$6 on average with a St. Dev of $5

What is the probability that on a given afternoon, the average of all 3 student's spending is more than $7?

I found that the chance of Student 1 spending more than $7 is .63056

Student 2's chance is .84134.

Student 3's chance is .42074

This is a total guess, but would it be the average of all 3?

(.63056+.84134+.42074)/3

Which would be 0.63088

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  • $\begingroup$ Hello RaiderNAYSHUN, welcome to cross validated. I think you got the steps needed right, but the order has to be changed. Right now you are calculating the avarage chance of each one of them spending more than 7\$, but the question is for the avarage spendings of all 3 beeing over 7\$. So i think you have to calc the distribution of the average of spendings of the 3 students first and after that you calc the chance for spendings > 7. At least thats how i would understand it. Maybe it brings the same result as your approach? Not sure about that :) $\endgroup$ – TinglTanglBob Aug 25 '18 at 21:31
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    $\begingroup$ You are implying student 3 will spend a negative amount 11.5% of the time. Could you provide a reasonable interpretation of that? You might also want to contemplate the (im)plausibility of the implicit assumption of independent spending amounts. $\endgroup$ – whuber Aug 25 '18 at 21:36
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    $\begingroup$ @TinglTanglBob Because "\$" is a $\TeX$ escape character on this site, you have to treat it specially. Preceding each \$ with a backslash "\" does the trick. $\endgroup$ – whuber Aug 25 '18 at 21:38
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    $\begingroup$ Hi: you need to find the distribution of the average of 3 independent normal random variables with different variances. Then, once you have that, you don';t need to worry about the fact that there are three people. You can just view it as a problem about a random variable. The trick is that the mean of the new random variable is a linear combination of the 3 means where the coefficients are 1/3 1/3 and 1/3. So, you can find the distribution and associated parameters of the new random variable using that relation. $\endgroup$ – mlofton Aug 25 '18 at 21:54
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    $\begingroup$ @whuber: Chance of finding a fiver while walking to the school cafeteria? (Good points of course. +1) $\endgroup$ – usεr11852 Aug 25 '18 at 22:21
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The specific problem may be flawed. But it is important to know that the distribution of the sum of independent normal random variables is normal. Here is a similar problem, but with very small probabilities of negative snack expenditures. Suppose we have independent random variables:

$X_1 \sim \mathsf{Norm}(\mu_1=9, \sigma_1=2),\,$ $X_2 \sim \mathsf{Norm}(\mu_2=10, \sigma_2=1),\,$ $X_3 \sim \mathsf{Norm}(\mu_3=12, \sigma_3=3).$

The probabilities of negative values for these three random variables are computed in R (where pnorm denotes a normal CDF) as follows:

pnorm(0, 9, 2);  pnorm(0, 10, 1);  pnorm(0, 12, 3)
[1] 3.397673e-06
[1] 7.619853e-24
[1] 3.167124e-05

It is OK to model test scores, human heights, and other distributions as normal, but you have to be careful. Even though most of the probability in a normal distribution lies within $\mu \pm 3\sigma,$ the values do extend down to $-\infty$ and up to $\infty.$ So you have to be sure that probabilities of impossible negative events are negligible. (That is the point of @whuber's first Comment.)

Then the distribution of the sum $S = X_1 + X_2 + X_3$ is $$\mathsf{Norm}\left(\mu_S = \mu_1+\mu_2+\mu_3,\; \sigma_S = \sqrt{\sigma_1^2+\sigma_2^2+\sigma_3^2}\right).$$

Then the average $A = S/3$ is distributed as $\mathsf{Norm}(\mu_S/3, \sigma_S/3).$

I suppose you can find these formulas in your text or class notes, and then find some probability such as $P(A < 10).$


Note: By simulating what happens on a very large number of afternoon snacks with these three students, it is possible to approximate $P(A < 10).$

m = 10^6;  x1 = rnorm(m, 9, 2);  x2 = rnorm(m, 10, 1);  x3 = rnorm(m, 12, 3)  
X = cbind(x1, x2, x3);  a = rowMeans(X)
mean(a);  sd(a);  sqrt(14)/3;  mean(a < 10)
[1] 10.33198  #  aprx E(A) = 31/3 
[1] 1.246459  #  aprx SD(A)
[1] 1.247219  #  exact SD(A)
[1] 0.395034  #  aprx P(A < 10)
pnorm(10, 31/3, sqrt(14)/3)
[1] 0.394634  #  exact P(A < 10)
mean(c(pnorm(10,9,2), pnorm(10,10,1), pnorm(10,12,3)))
[1] 0.4813183 #  Your method of averaging probabilities doesn't work

The vector a contains a million average snack bills; a < 10 is a logical vector with a million TRUEs and FALSEs. Its mean is the proportion of its TRUEs.

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