Question

If a die is rolled 300 times, use the Chebyshev inequality to estimate the probability
that the number of occurrences of "three" does not lie strictly between 45 and 55.

Answer 1

Probability that "three" occurs on a random toss of a fair die = 1/6

Let X be a Binomial random variable which denotes the number of die in which "three" occurs on 300 tosses of a die

Mean of X = np = 300*1/6 = 50

Standard deviation of X =

= 6.455

Since both np and n(1 - p) are greater than 10, X can be approximated to Normal distribution with mean = 50 and standard deviation = 6.455

Probability that the number of occurrences of "three" does not lie between 45 and 55 = 1 - P(45 ≤ X ≤ 55)

= P(X < 45) + P(X > 55)

Using correction of continuity (since the normal approximation is done from a discrete random variable), the required probability = P(X < 44.5) + P(X > 55.5)

= P(|X - | > 5.5)

= P(|X - | > 0.852)

Using Chebyshev's inequality,

P(|X - | > 0.852) ≤ ≤ 1.378

which is obvious because the probability cannot be greater than 1

Using z table, the required probability = 0.6057