Question

Suppose the heights of students are normally distributed with mean 172 and variance 9.

a) Randomly pick two students. What's the probability that the first student is taller than the second student?

b) Randomly pick two students. What's the probability their average height is larger than 175?

Answer 1

Heights of students are normally distributed with mean 172 and variance 9

Thus X~ N(172 , 9 )

a) Randomly pick two students, to find probability that the first student is taller than the second student.

Now we don't have size neither have any idea about Randomly picked two students .

So we can do following things

Let us define two variables x and y correspond to two students .

Generate sample of size n = 100,500,1000 form X ~ N( 172 , 9 )

Choose random samples from generated sample of normal distribution of size n/2 for both variables

Now compare now variables.

We will use R Software

Command-

{

n=100                              # sample size
r=rnorm(n,172,3)                               # genrate samples from N( = 172, = 3)

for (i in 1:50)                                     # we will take 50 repitaions
{
x=sample(r,n/2)                                          # first student randomly selected from normal distribution
y=sample(r,n/2)                                         # second student randomly selected from normal distribution

# probability that the first student is taller than the second student   

# P(x>y)
prob[i]=length(which(x>y))/length(x)          # probability of first student taller than the second student
}

OUTPUT -

> prob
[1] 0.44 0.36 0.54 0.44 0.56 0.44 0.44 0.40 0.50 0.44 0.60 0.54 0.54 0.60 0.52
[16] 0.56 0.46 0.52 0.50 0.50 0.52 0.54 0.46 0.46 0.48 0.56 0.58 0.46 0.46 0.50
[31] 0.46 0.50 0.46 0.46 0.48 0.50 0.46 0.52 0.54 0.54 0.44 0.42 0.52 0.48 0.52
[46] 0.40 0.50 0.60 0.52 0.50

> mean(prob)
[1] 0.4948

Now we will repeat same program for n= 1000

n=1000
r=rnorm(n,172,3)

for (i in 1:50)
{
x=sample(r,n/2)
y=sample(r,n/2)
prob[i]=length(which(x>y))/length(x)
}

OUTPUT -

> prob
[1] 0.502 0.480 0.476 0.504 0.472 0.492 0.514 0.476 0.518 0.488 0.504 0.500
[13] 0.524 0.520 0.502 0.538 0.504 0.506 0.484 0.478 0.486 0.512 0.488 0.490
[25] 0.508 0.482 0.504 0.502 0.522 0.526 0.516 0.502 0.520 0.508 0.514 0.486
[37] 0.538 0.538 0.538 0.512 0.504 0.482 0.522 0.494 0.482 0.472 0.504 0.478
[49] 0.522 0.496

> mean(prob)
[1] 0.5026

Conclusion

Form both output we can conclude that probability that the first student is taller than the second student

is approximately equal to 0.50

b)

Randomly pick two students. To find probability that their average height is larger than 175

i.e   ( x + y )/2 > 175

Now Let T = ( x + y )/2

Given

= 172, = 3

To find P( T>175)

Now from CLT

Z = ( T - 172 ) / n^1/2

But we don't have n

So we can use similar process used in part (a) now to find T > 175

Command

{

n =1000                                                     # sample size
r=rnorm(n,172,3)

for (i in 1:500)                                                         # 500 iterations
{
x=sample(r,n/2)
y=sample(r,n/2)
avg=(x+y)/2                                               #   avg   =    T = ( x + y ) / 2
prob[i]=length( which( avg>175 ))/length(x)            # P( T>175)
}

mean(prob)

}

Output -

mean(prob)                # for n = 1000
[1] 0.079768

mean(prob)                 # for n = 1500

[1] 0.083408

mean(prob)                # for n = 1750       
[1] 0.08051429

mean(prob)                 # for n = 2000
[1] 0.082236

Conclusion probability their average height is larger than 175 is approximately = 0.08