1. If College Board (CEEB) scores are normally distributedwith a mean of 7OO and

Question

1. If College Board (CEEB) scores are normally distributedwith a mean of 7OO and a standard deviation of 1OO;
a) . what percent of students taking the exam have a score of 75Oor above?

b). what score represents P90 (ninetiethpercentile)?

2.    A certain teacher has discovered that over theyears the average grade on a 150-point comprehensive exam is 110with a standard deviation of 12. She has decided that onlythe top 5% get an A on the exam; the next 15% a B the next 40% a C.What score and corresponding percentage
( i e. score/150 must a student get to receive an A? a B?.

Explanation / Answer

1. a) First, standardize the normal distribution so we can use thetable. P(X > 750) = 1 - P(X < 750) = 1 - P(Z < (750-700)/100) = 1- P(Z < 0.5) Here is a sample table:http://business.statistics.sweb.cz/normal01.jpg According to the table, P(Z < 0.5) = 0.6915. So the percentageof students with scores of 750 and above is 1 - 0.6915 =0.3085. b) Here, we first find the number on the standard normal Z(0,1) tothe left of which is 90% of the probability, and it is 1.28. Nowun-normalize the distribution: 1.28 = (X - 700)/100 X = 828 P90 is thus 828. 2. The distribution of the exam is N(110, 144). We first find thenumber on the standard normal to the left of which is 95% of theprobability, and it is 1.645. Similar to the last question, weun-normalize the distribution: 1.645 = (X - 110)/12 X = 129.7 129.7 is what a student would need to receive an A. For the second part of the question, we need the number on thestandard normal to the left of which is 80% of the probability, andit is 0.84. In an analogous fashion: 0.84 = (X - 110)/12 X = 120.1 120.1 is what a student would need to receive a B.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.