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 1180794: 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?
what is the area under the normal distribution curve between z= -1.3 and z= 0.99?
for a specific confidence interval,what will happen to the margin of error if the sample size is larger?
a.it becomes smaller
b.it becomes larger
c. equal
d. the same
when a 99% confidence interval is calculated instead of a 95% confidence interval with n being the same, the margin of error will be.
a.smaller
b.larger
c. the same
d. it cannot be determine
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)

ME+=+z%2Asigma%2Fsqrt%28n%29
 As n gets larger...ME gets smaller
 As z gets larger(n the same) ...ME gets larger
 = CI	z = value
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576

Important You feel comfortable using Your calculator.
Wish You the Best in your Studies.