Consider the random variable X ∼ N (5, 9); that is, a normal random variable with mean µ = 5and standard deviation σ = 3.(a) Evaluate the following:P(X ≤ 2), P(X > 2), P(X ≤ 8), P(X > 8).(b) Find the points x such that:P(X ≤ x) = 0.10, P(X ≤ x) = 0.25, P(X ≤ x) = 0.52, P(X ≤ x) = 0.99.Page 2 Page 2 of 3

(a) To evaluate the probabilities, we need to standardize the values using the Z-score formula: Z = (X - µ) / σ.

1. P(X ≤ 2): This is equivalent to finding the probability that Z ≤ (2-5)/3 = -1. So, we need to find P(Z ≤ -1). Using the standard normal distribution table, we find that P(Z ≤ -1) = 0.1587.

2. P(X > 2): This is equivalent to 1 - P(X ≤ 2) = 1 - 0.1587 = 0.8413.

3. P(X ≤ 8): This is equivalent to finding the probability that Z ≤ (8-5)/3 = 1. So, we need to find P(Z ≤ 1). Using the standard normal distribution table, we find that P(Z ≤ 1) = 0.8413.

4. P(X > 8): This is equivalent to 1 - P(X ≤ 8) = 1 - 0.8413 = 0.1587.

(b) To find the points x such that P(X ≤ x) = a certain value, we need to use the inverse of the standard normal distribution function (also known as the quantile function or the percent-point function).

1. P(X ≤ x) = 0.10: Using the inverse of the standard normal distribution function, we find that Z = -1.28. Then, we use the formula X = µ + Zσ to find x = 5 + (-1.28)(3) = 1.16.

2. P(X ≤ x) = 0.25: Using the inverse of the standard normal distribution function, we find that Z = -0.67. Then, we use the formula X = µ + Zσ to find x = 5 + (-0.67)(3) = 3.01.

3. P(X ≤ x) = 0.52: Using the inverse of the standard normal distribution function, we find that Z = 0.05. Then, we use the formula X = µ + Zσ to find x = 5 + (0.05)(3) = 5.15.

4. P(X ≤ x) = 0.99: Using the inverse of the standard normal distribution function, we find that Z = 2.33. Then, we use the formula X = µ + Zσ to find x = 5 + (2.33)(3) = 11.99.