The marks of 500 candidates in an examination are normally distributed
with mean of 45 marks and a standard deviation of 20 marks.
a) Given that the pass mark is 41, estimate the number of candidates
who passed the examination.
b) If 5% of the candidates obtained a distinction by scoring x marks or
more, estimate the value of x.
Let "Y=" the mark in an examination: "Y\\sim N(\\mu, \\sigma^2)"
Given "\\mu=45, \\sigma=20."
a)
"=1-P(Z<\\dfrac{41-45}{20})=1-P(Z<-0.2)"
"\\approx0.579260"
"500(0.579260)=289.63"
289 candidates passed the examination.
b)
"=1-P(Z<\\dfrac{x-45}{20})=0.05"
"P(Z<\\dfrac{x-45}{20})=0.95"
"\\dfrac{x-45}{20}=1.6449"
"x=78"
Comments
Leave a comment