Suppose X is a normally distributed random variable with unknown mean μ and unknown standard deviation σ. Assume a random sample of X values, of size n = 25, is available. Let s^2 denote the sample variance. If σ^2=40, what is P(S^2≤23.08)?

Question

Suppose X is a normally distributed random variable with unknown mean μ and unknown standard deviation σ. Assume a random sample of X values, of size n = 25, is available. Let s^2 denote the sample variance.  If σ^2=40, what is P(S^2≤23.08)?

john_es · Answer

$$P(S^2\leq23.08)=P(\frac{nS^2}{\sigma^2}\leq\frac{25\cdot23.08}{40})$$
$$P(\chi^2_{24}\leq14.43)=0.6$$
You must do an interpolation in order to get the value, or an online calculator.