SOLUTION: THE SCORES OF AN ENTRANCE TEST FOR HIGH SCHOOL PASS-OUTS IN A PARTICULAR YEAR WERE BELL-SHAPED. THE MEAN AND STANDARD DEVIATION WERE 490 AND 100 RESPECTIVELY. WHAT PERCENTAGE OF

Algebra ->  Probability-and-statistics -> SOLUTION: THE SCORES OF AN ENTRANCE TEST FOR HIGH SCHOOL PASS-OUTS IN A PARTICULAR YEAR WERE BELL-SHAPED. THE MEAN AND STANDARD DEVIATION WERE 490 AND 100 RESPECTIVELY. WHAT PERCENTAGE OF      Log On


   

Question 1180790: THE SCORES OF AN ENTRANCE TEST FOR HIGH SCHOOL PASS-OUTS IN A PARTICULAR YEAR WERE BELL-SHAPED. THE MEAN AND STANDARD DEVIATION WERE 490 AND 100 RESPECTIVELY.
WHAT PERCENTAGE OF THE STUDENTS SCORED BETWEEN 390 AND 590 ON THIS TEST?
A.2.35% B.13.5% C.34% D.68%
WHAT IS THE AREA UNDER THE NORMAL DISTRIBUTION CURVE BETWEEN Z= -1.3 AND Z=0.99?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
Normal Distribution:  µ = 490 and σ = 100 
Using TI or similarly an inexpensive calculator like a Casio fx-115 ES plus
Continuous curve

P(390 > x > 590)= normalcdf(390, 590, 490, 100) = .6827  0r 68% (2 decimal places)
Keisan Online Calculator
   .

P(-1.3 > z > 8.99)= normalcdf(-1.3, 8.99, 0, 1) = .90 (2 decimal places)

Important You feel comfortable using Your calculator.
Wish You the Best in your Studies.
standard Normal Curve:  µ = 0 and σ = 1
various z-values given