Question

The average height for a student in a class with n = 60 students is a random variable with an average height of 180 cm and standard deviation σ = 10. The individual heights which make up that average are i.i.d. (1) use Chebyshev’s inequality to find an upper bound for the probability that the average of the class (obtained from the individual student heights) is greater than 200cm. (2) Use Chebyshev’s inequality to upper bound the probability that the average height of a given student is above 195 centimeters. Additionally, (3) Use the CLT to approximate the probability that the monthly average (found from the average of each individual student height) is greater than 200cm.

Answer 1

Let X be the random variable which denotes average height of each student in a class of n= 60 students with mean= Average height(X) =E(X) = 180cm and Standard deviation(X)=S(X) =10cm and variance V(X) =102 =100

Then Average of the class = sample mean = xbar is also random variable with mean

E(xbar) = E(X) = 180 and V(xbar) = V(X)/n =100/60 =5/3

We know the Chebysheve’s inequality as

P(lX-E(X)l>=a)<=V(X)/a2 where, a>0

P(lX-180l>=a)<=100/a2 from the given data

P(lXl>=a+180)<=100/a2

P(X>=a+180)<=100/a2 ….(1)

Also, the chebysheve’s inequality is P(lxbar-E(xbar)l>=a)<=V(xbar)/a2 where, a>0

P(lxbar-180l>=a)<=5/(3a2 ) from the given data

P(lxbarl>=a+180)<=5/(3a2 )

P(xbar>=a+180)<=5/(3a2 )    ….(2)

(1) By using the above equation (2) of Chebyshev’s inequality to find an upper bound for the probability that the average of the class (obtained from the individual student heights) is greater than 200cm, we obtain

P(xbar>200)

= P(xbar>20+180) <=5/(3*202 )    (from(2))

<=5/(3*400 )

<=1/(3*80 )

<=1/(240 )

<=0.0042

(2) By using the above equation (1) of Chebyshev’s inequality to find an upper bound for the probability that the average height of a given student is above 195 centimeters, we obtain

P(X>195)

= P(X>15+180) <=100/152   (from (1))

<=100/225

<=4/9

<=0.4444

(3) According to central limit theorem(CLT) mean of n independent observations follows normal distribution with mean as E (mean of n independent observations) and variance as V(mean of n independent observations) if n is large preferably n>=30.

Since, here n=60 (n>30) then the monthly average (found from the average of each individual student height) that is the class average sample mean xbar follows normal distribution with mean E(xbar) =E(X) =180 and variance v(xbar) = V(X)/n = 100/60 = 5/3 according to CLT.

Then standard normal variable is z = (xbar- E(xbar))/sqrt(V(xbar))

z = (xbar- 180)/sqrt(5/3) = (xbar- 180)/1.29

By using the CLT to approximate the probability that the monthly average (found from the average of each individual student height) is greater than 200cm, we obtain

P(xbar>200) = P(z>(200-180)/1.29)

= P(z>20/1.29)= P(z>15.5)= almost equal to 0.