- #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.

$$Z = \frac {40- 50} {15} = -0.67$$

$$N(-0.67) = 1- N(0.67) $$

$$1- N(0.67) = 1-0.7486 = 0.2514$$

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.