Question

5. North Carolina State University posts the complete grade distributions for its courses online. The distribution of grades for all students in all sections of Accounting 210 in the spring semester of 2001 was Grade Probability .18 32 34 09 07 a. Using the scale A -4, B-3, C-2, D- 1, and F 0, let Xbe the grade of a randomly chosen b. Let X denote the mean grade for a random sample of 50 students from Accounting 210. Since and standard deviation ơ of Accounting 210 student. Compute the mean there are a finite number of students in the class each semester, sampling from this finite population actually results in slight dependence within the sampling. However, an extremely large number of students take Accounting 201 each semester (much larger than the size of our sample), so it is reasonable to treat the grades as approximately independent. Assuming independence, what does the Central Limit Theorem say about the probability distribution of X? What are the mean and standard deviation of X? c. What is the probability P(X 23) that a randomly selected Accounting 210 student gets a B or better? What is the approximate probability P(X23) that the average grade for 50 randomly chosen Accounting 210 students is B or better?

Answer 1

a)

Σ r. P(x)

= 4*0.18 +......+0*0.07

, = 2.45

$\sigma ^{2} = \sum x^{2}P(x) - (\sum xP(x)))^{2}$

=1.2075

therefore ,

$\sigma =1.0989$

b) The sampling distribution of sample mean, $\bar{X}$ follow normal distribution

with mean , (population mean )

E(X) = μ

E(X) = 2.45

and standard deviation of $\bar{X}$ , known as standard error , $\sigma _{\bar{X}}= \sigma /\sqrt{n}$ = 0.1554

where $\sigma$ is the population standard deviation and n is the sample size .

c)

P(X > 3) = 0.32 + 0.18 = 0.50

with mean = 2.45 and standard error = 0.1554

Y Normal

then $Z= \frac{\bar{X}-2.45 }{0.1554}\sim N(0,1)$

P(X 23)-P: 23.54)

= 0.0002