Question

point) The professor of a Introductory Calculus class has stated that, historically, the distribution of final exam grades in the course resemble a Normal distribution with a mean final exam mark of 63% and a standard deviation of = 11% using/Tinding z-values, use three decimals (a) What is the probability that a random chosen final exam mark in this course will be at least 75%7 Answer to four decimals. 0.1378 a) in order to pass this course, a student must have a final exam mark of at least 50%. What proportion of students will not pass the calculus final exam? Use four decimals in your answer 0.8814 (c) The top 4% of students writing the final exam will receive a better grade of at least an A in the course. To two decimal places, find the minimum final exam mark needed on the calculus final to earn a letter grade of at least an A in the course. id) Suppose this professor randomly picked 28 final exams, observing the earned mark on each. What is the probability that of these have a final exam grade of less than 50X7 Une four decimals in your answer.

Answer 1

Solution from the given information, Here, Mean l = 63% std (6) - 11 % X ~Normal = 2 = x-1 - 63 a Plat least 75%) = pla => Plz> 1.0909) - 1-P (2<1.091) -) 1-0.8623 -0.1377) ② plat least 50%) = P(x250) will not pass plx (50 - Plac 30-03) - P(2<- 1.182) 0-1186

0 Top 4%. Pla<x) = 0.96. Plxxx) = 0.04 we know 96th percentile value for 2 is 1.75) minimum marts to get A= 63+ 1.75 X 11 (A = 82.25% Here, use Binomial distribution with n=28 and p=0.1186 22 p(6 have less than 50%) - 286 +0.1186 + (1 - 011 86² >10.06521
Please rate if it is really helps you. Thank you.