A random variable X has a normal probability distribution N (0, 1). De-termine the value of c as a function of u and o such thatP (X ≥ c) = 9P (X < c).(a) Find the value of c.(b) Evaluate P ((X − µ)2 ≥ σ)

Sure, let's solve this step by step.

(a) We are given that P(X ≥ c) = 9P(X < c). This implies that the probability of X being greater than or equal to c is 9 times the probability of X being less than c. Since the total probability must be 1, we can write this as:

1 = P(X ≥ c) + P(X < c) = 9P(X < c) + P(X < c) = 10P(X < c)

Solving for P(X < c) gives P(X < c) = 1/10.

Since X follows a standard normal distribution, we can use the standard normal distribution table or the z-score formula to find the value of c. The z-score corresponding to a cumulative probability of 0.1 (or 1/10) is approximately -1.28.

Therefore, c = -1.28.

(b) The expression P((X − µ)2 ≥ σ) is the probability that the squared deviation of X from its mean µ is greater than or equal to the standard deviation σ.

Since X follows a standard normal distribution, its mean µ = 0 and standard deviation σ = 1. Therefore, the expression simplifies to P(X^2 ≥ 1).

The square of a standard normal random variable follows a chi-square distribution with 1 degree of freedom. The cumulative distribution function of a chi-square distribution with 1 degree of freedom at x = 1 is approximately 0.6827.

Therefore, P(X^2 ≥ 1) = 1 - P(X^2 < 1) = 1 - 0.6827 = 0.3173.