Average of 3 normally distributed random samples

In summary: 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).
  • #1
TheBigDig
65
2
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.
 
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  • #2
Hi there,

TheBigDig said:
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
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
Master1022 said:
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?

pasmith said:
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
TheBigDig said:
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
 

1. What is the definition of a "normally distributed random sample"?

A normally distributed random sample is a set of data points that follow a normal distribution, also known as a Gaussian distribution. This means that the data is symmetrically distributed around a central mean value, with the majority of the data falling within a certain range of the mean.

2. How is the average of 3 normally distributed random samples calculated?

The average of 3 normally distributed random samples is calculated by adding together all of the data points from each sample and dividing by the total number of data points. This is also known as the arithmetic mean.

3. What is the significance of using 3 samples instead of just 1?

Using 3 samples instead of just 1 allows for a more accurate representation of the data. This is because having multiple samples helps to reduce the impact of outliers and random variations, resulting in a more reliable average.

4. How does the shape of the normal distribution affect the average of 3 samples?

The shape of the normal distribution does not significantly affect the average of 3 samples as long as the data is normally distributed. This is because the average is calculated by taking into account all of the data points and their relative positions, rather than just the shape of the distribution.

5. Can the average of 3 normally distributed random samples be used to make predictions?

Yes, the average of 3 normally distributed random samples can be used to make predictions, as long as the data is collected and analyzed properly. However, it is important to note that the accuracy of the predictions will depend on the quality and representativeness of the data.

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