Let S2 denote the sample variance from a random sample of si

Let S² denote the sample variance from a random sample of size n = 25 from a normal distribution with variance ² = 4.

(a) Find the probability that S² > 5.53.

(b) For large v we may regard a ² random variable with v degrees of freedom as an approximately normal random variable with mean v and variance 2v. Use normal approximation to instead estimate P(S² > 5.53)

Solution

We are given that, S2 denote the sample variance.

sample size is 25.

variance 2 = 4.

Find the probability that S2 > 5.53.

P(S2 > 5.53) = P(n*S2 / 2 > n*5.53/ 2 )

= P(X² > (25*5.53) / 4 )

= P(X² > 34.5625)

=1 - 0.075

= 0.925

For large v we may regard a 2 random variable with v degrees of freedom as an approximately normal random variable with mean v and variance 2v. Use normal approximation to instead estimate P(S2 > 5.53)

mean (v) = 24

variance (2v) = 2*24 = 48

P((S2 - mean) / sd > (5.53 - 24)/sqrt(48))) = P(Z > -2.6659)

P(Z > -2.6659) = 1 - P(Z <=-2.6659)

We can find this probability by using EXCEL.

syntax is,

=NORMSDIST(z)

P(Z <=-2.6659) = 0.0038

P(Z > -2.6659) = 1 - 0.0038 = 0.9962