Question

1. A group of students with normally distributed salaries earn an average of $6,800 with a standard deviation of $2,500. Use Excel Formulae/functions to answer.

(a) What proportion of students earn between $6,500 and $7,300?

(b) What are the first and third quartiles of students’ salaries?

(c) What value of salary in $ exceeded the 95% probability?

Answer 1

Let x be the salaries of student's. XN NC 6,800 (2500)2 ). a) proportion of student's earn between $6500 & $7500 = P (6500L X 27300) - pl y<7300) - PCX66500) Using Ms. Excel PCX (7300) - NORN OLST ( 7300, 6800, 2500, 1) Toc XC6500) = NORM DIST (6500, 6800, 500, 1) -) p (6500CX (7300 ) = 0.5792 -0.4522 -0.1270 Proportion of student's eam between $6500 & 7300 is 0.1270 b) 1st quantileg is the values such that ast. shof salaries are less than gi ire. PCXC91) = 0.25 = 0, 2 0 (0:25) where is codofi of normal distribution * Using Ms Excel 110: 25) = NORMINU (0.25, 6800 ,2500) =) jst Quartile = 5113. 77 5625. grd quantilegis such that P ( X (93) = 0.75 3 93 = 0 (0.75) . Using ths - Excel 0-1 (0.7) = NORMINU (0.75,6800, 2500) =8486.22 4875 = 3rd quantile=8486.224375

c) Salary is $ excaeeded gsat. probability is given by Xt such that PCX<X ) zogs '* x* ? 0-1 (0.95). Using MS-Excel 0-1 (0.95) - NORM INV(o-95,6800,2500) 30 (0.95) = 10.9 12.13407 This is the required salary.