Question

In an examination the probability distribution of scores (X) can be approximated by normal distribution with mean 63.2 and standard deviation 7.6.

*Please explain how you got your answers.

1) Suppose one has to obtain at least 55 to pass the exam. What is the probability that a randomly selected student passed the exam? [Answer to 3 decimal places]

Tries 0/5

2) If two students are selected randomly what is the chance that both the students failed? [Answer to 3 decimal places]

Tries 0/5

3) If only top 8% students are given an award, then what was the minimum marks required to get the award? [Answer to 1 decimal place]

Answer 1

a)
mean= 63.20
sd= 7.60

P( X>55)
= 1 - P(X<55)

I know that, z = (X-mean)/(sd)
z1 = (55-63.2)/7.6)
z1= -1.0789

hence,
P( X>55)
=1- P(Z<-1.0789)
1 - NORMSDIST(-1.0789)
0.860

b)
mean= 63.2
sd= 7.6
n= 2

P(X < 55)
I know that, z = (X-mean)/(sd/sqrt(n))
z1 = (55-63.2)/7.6/sqrt(2))
z1= -1.5259

hence,
P(X < 55)=
= P(Z<-1.5259)
NORMSDIST(-1.5259)
0.064

c)
P(Z<z) 92%
z= NORMSINV(0.92)
z= 1.40507156
I know that, z = (X-mean)/sd
(X-mean)/sd = 1.4051
X= 1.4051*7.6+63.2
X= 73.9