Question

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 A B C D F

Probability 0.18 0.32 0.34 0.09 0.07

a. Using the scale A = 4, B = 3, C = 2, D = 1, and F = 0, let X be the grade of a randomly chosen Accounting 210 student. Compute the mean μ and standard deviation σ of X.

b. Let X denote the mean grade for a random sample of 50 students from Accounting 210. Since 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 ≥ 3) that a randomly selected
Accounting 210 student gets a B or

better? What is the approximate probability P ( X ≥ 3) that the average grade for 50 randomly chosen Accounting 210 students is B or better?

Answer #1

(a) mean= μ=E(x)=sum(x*p(x))=2.45

E(x^{2})=sum(x^{2}*p(x))=7.21

variance(x)=E(x2)-E(x)E(x)=7.21-2.45*2.45=1.2075

standard deviation =σ =sqrt(variance(x))=sqrt(1.2075)=1.0988

A | B | C | D | F | sum | |

x | 4 | 3 | 2 | 1 | 0 | 10 |

p(x) | 0.18 | 0.32 | 0.34 | 0.09 | 0.07 | 1 |

x*p(x) | 0.72 | 0.96 | 0.68 | 0.09 | 0 | 2.45 |

x2*p(x) | 2.88 | 2.88 | 1.36 | 0.09 | 0 | 7.21 |

(b) the mean=2.45 and standard deviation =1.9088

(c) here we use standard normal variate z=(x-mean)/(sd/sqrt(n))

for x=3, z=(3-2.45)/(1.9088/sqrt(50))=2.0375

P(X>=3)=1-P(X<3)=1-P(Z<2.0375)=1-0.9792=0.0208

required probability=0.0208

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