Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The scores of students in a national level examination are normally distributed with a mean of 500 and a standard deviation of 100. If the value of the cumulative distribution of the standard normal random variable at 0.5 is 0.691, then the probability that a randomly selected student scored between 450 and 500 is
Solution
Given:Scores are normally distributed with
mean $ \mu = 500 $,
standard deviation $ \sigma = 100 $
We want the probability that a student scored between 450 and 500:
$P(450 < X < 500)$
Convert to $Z$-scores:
For $X = 450$:
$Z = \dfrac{450 - 500}{100} = -0.5$
For $X = 500$:
$Z = \dfrac{500 - 500}{100} = 0$
Thus:
$P(450 < X < 500) = P(-0.5 < Z < 0)$
Given:
$P(Z < 0.5) = 0.691$
Using symmetry of the standard normal curve:
$P(-0.5 < Z < 0.5) = 2(0.691 - 0.5) = 2(0.191) = 0.382$
So:
$P(-0.5 < Z < 0) = \dfrac{0.382}{2} = 0.191$
Final answer: 0.191
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
