Average of 3 normally distributed random samples

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Homework Statement:
A new university teacher of a large first year class is uncertained how to distribute grades. She decides to assume that the scores will be normally distributed with mean ##\mu## and standard deviation ##\sigma##. There are 5 grades (A,B,C,D,E) with A being the highest.

(i) Suppose that the scores are normally distributed with mean 50 and standard deviation 15. What percentage of students will fail if the pass mark is 40%?

(ii) If a random sample of three scripts is selected by the external examiner for detailed scrutiny, what is the probability that the average score for these scripts will be less than 40?

(iii) Explain why it would be surprising if the sum of the 3 scores exceeds 250.
Relevant Equations:
$$Z = \frac {x- \mu} {\sigma},$$
I've found part (i) by calculating the z-score for 40
$$Z = \frac {40- 50} {15} = -0.67$$
$$N(-0.67) = 1- N(0.67) $$
$$1- N(0.67) = 1-0.7486 = 0.2514$$

But parts (ii) and (iii) are confusing me. I have answers provided by my professor that say the mean and std deviation for (ii) and (iii) are
(ii) ##\mu = 50, \sigma = 15/\sqrt(3)##
(iii) ##\mu = 150, \sigma = 15\sqrt(3)##
but I'm not sure where the standard deviations come from as it's not something we've covered in class.
 

Answers and Replies

  • #2
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Hi there,

But parts (ii) and (iii) are confusing me. I have answers provided by my professor that say the mean and std deviation for (ii) and (iii) are
(ii) ##\mu = 50, \sigma = 15/\sqrt(3)##
(iii) ##\mu = 150, \sigma = 15\sqrt(3)##
but I'm not sure where the standard deviations come from as it's not something we've covered in class.

So these results come from the combination of independent random variables. We realize that we need to find the parameters for the new normal distributions.

For part (ii), we are picking three tests at random [itex] X_1 [/itex], [itex] X_2 [/itex], and [itex] X_3 [/itex]. We are taking the scores and averaging them so we can define a new variable [itex] Y [/itex] to be our average score
[tex] Y = \frac{X_1 + X_2 + X_3}{3} [/tex]
Now if we want to find the mean (expected value) and variance of our new distribution, then we can use the fact that (where [itex] a [/itex] and [itex] b [/itex] are constants)
[tex] E[aX + b] = aE[X] + b [/tex]
and
[tex] Var(aX + b) = a^2 Var(X) [/tex]
Therefore, given that [itex] E[X_i] = 50 [/itex] and [itex] Var(X) = 15 [/itex]:
[tex] E[Y] = \frac{X_1 + X_2 + X_3}{3} = \frac{1}{3} E[X_1] + \frac{1}{3} E[X_2] + \frac{1}{3} E[X_3] = 50 [/tex]
and
[tex] Var(Y) = \left( \frac{1}{3} \right)^2 \left( Var(X_1) + Var(X_2) + Var(X_3) \right) = \frac{15^2}{3} [/tex]
Therefore, the standard deviation is: [itex] \sigma_Y = \sqrt{Var(Y)} = \frac{15}{\sqrt{3}} [/itex]

For part(iii), we now don't want the average of the three scores, but instead just the sum. So we can define the random variable [itex] Z [/itex] as:
[tex] Z = X_1 + X_2 + X_3 [/tex]
Using similar results to above, the standard deviation can be derived. Let me know if you aren't able to get to the result. Please note that we are adding the three test scores rather than multiplying by 3 - there is a subtle difference when it comes to the resultant variance (can discuss this below, but don't want to confuse you unnecessarily).

These are results from combining random variables and align with results from the 'Central Limit Theorem', which I would recommend looking up (there is plenty of material out there) and can be generalized for a general number [itex] n [/itex] of distributions combined.

Hope that was of some help. There are some general formulae for combining normal distributions
 
  • #3
pasmith
Homework Helper
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778
Did you not cover the facts that if [itex]X_1[/itex] and [itex]X_2[/itex] are independent and normally distributed with means [itex]\mu_1[/itex] and [itex]\mu_2[/itex] and variances [itex]\sigma_1^2[/itex] and [itex]\sigma_2^2[/itex] then
  • [itex]X_1 + X_2[/itex] is normally distributed with mean [itex]\mu_1 + \mu_2[/itex] and mean [itex]\sigma_1^2 + \sigma_2^2[/itex], and
  • [itex]\alpha X_1[/itex] is normally distributed with mean [itex]\alpha\mu_1[/itex] and variance [itex]\alpha^2 \sigma_1^2[/itex]
and the consequence that if a sample of size [itex]n[/itex] is taken from a normal distribution with mean [itex]\mu[/itex] and variance [itex]\sigma^2[/itex] then the sample mean is normally distributed with mean [itex]\mu[/itex] and variance [itex]\sigma^2/n[/itex]?
 
  • #4
65
2
Hi there,



So these results come from the combination of independent random variables. We realize that we need to find the parameters for the new normal distributions.

For part (ii), we are picking three tests at random [itex] X_1 [/itex], [itex] X_2 [/itex], and [itex] X_3 [/itex]. We are taking the scores and averaging them so we can define a new variable [itex] Y [/itex] to be our average score
[tex] Y = \frac{X_1 + X_2 + X_3}{3} [/tex]
Now if we want to find the mean (expected value) and variance of our new distribution, then we can use the fact that (where [itex] a [/itex] and [itex] b [/itex] are constants)
[tex] E[aX + b] = aE[X] + b [/tex]
and
[tex] Var(aX + b) = a^2 Var(X) [/tex]
Therefore, given that [itex] E[X_i] = 50 [/itex] and [itex] Var(X) = 15 [/itex]:
[tex] E[Y] = \frac{X_1 + X_2 + X_3}{3} = \frac{1}{3} E[X_1] + \frac{1}{3} E[X_2] + \frac{1}{3} E[X_3] = 50 [/tex]
and
[tex] Var(Y) = \left( \frac{1}{3} \right)^2 \left( Var(X_1) + Var(X_2) + Var(X_3) \right) = \frac{15^2}{3} [/tex]
Therefore, the standard deviation is: [itex] \sigma_Y = \sqrt{Var(Y)} = \frac{15}{\sqrt{3}} [/itex]

For part(iii), we now don't want the average of the three scores, but instead just the sum. So we can define the random variable [itex] Z [/itex] as:
[tex] Z = X_1 + X_2 + X_3 [/tex]
Using similar results to above, the standard deviation can be derived. Let me know if you aren't able to get to the result. Please note that we are adding the three test scores rather than multiplying by 3 - there is a subtle difference when it comes to the resultant variance (can discuss this below, but don't want to confuse you unnecessarily).

These are results from combining random variables and align with results from the 'Central Limit Theorem', which I would recommend looking up (there is plenty of material out there) and can be generalized for a general number [itex] n [/itex] of distributions combined.

Hope that was of some help. There are some general formulae for combining normal distributions

Hi there! Thank you so much for your help. As it turns out we have covered this before but I didn't consider that these manipulations could be performed when it was three variables taken from the same data set if you understand me. I thought it only applied to normal variables with different means and variances but I see now why it makes sense. I got the standard deviation in part (iii) but I looked into it and see what you mean about it being more nuanced than just multiplying them by three

$$var(aX + bY +cZ) = a^2Var(X)+b^2Var(Y)+c^2Var(Z)+2abCov(X,Y)+2acCov(X,Z)+2bcCov(Y,Z)$$

I assume in this example since they all come from the dataset we can take covariance to be zero?

Did you not cover the facts that if [itex]X_1[/itex] and [itex]X_2[/itex] are independent and normally distributed with means [itex]\mu_1[/itex] and [itex]\mu_2[/itex] and variances [itex]\sigma_1^2[/itex] and [itex]\sigma_2^2[/itex] then
  • [itex]X_1 + X_2[/itex] is normally distributed with mean [itex]\mu_1 + \mu_2[/itex] and mean [itex]\sigma_1^2 + \sigma_2^2[/itex], and
  • [itex]\alpha X_1[/itex] is normally distributed with mean [itex]\alpha\mu_1[/itex] and variance [itex]\alpha^2 \sigma_1^2[/itex]
and the consequence that if a sample of size [itex]n[/itex] is taken from a normal distribution with mean [itex]\mu[/itex] and variance [itex]\sigma^2[/itex] then the sample mean is normally distributed with mean [itex]\mu[/itex] and variance [itex]\sigma^2/n[/itex]?

Thank you as well for your help!
 
  • #5
582
111
I assume in this example since they all come from the dataset we can take covariance to be zero?

Yes, we can assume they are independent so that the covariance is 0
 

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