Question

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

Short answer

a. 0.9906, b. 0.9906, c. 0.1093, d. 0.0101, e. 0.0588, f. 0.0228, g. 0.9996, h. 1

Step-by-step solution

Understanding Standard Normal Distribution Table

A standard normal distribution table, or Z-table, is used to find the probability that a statistic is less than a given Z-score. This table gives the area (probability) that lies to the left of a given Z-score.

Finding the probabilities

Using the Z-table, find the probabilities for each part of the exercise. Here are the probabilities for each part: a. \(P(z<2.36)\) = 0.9906b. \(P(z \leq 2.36)\) = 0.9906 (Whether the inequality is less than or less than and equal to does not matter in this context because we're dealing with a continuous distribution.)c. \(P(z<-1.23)\) = 0.1093d. \(P(1.142)\) = 1 - 0.9772 = 0.0228 (for z-scores greater than a number, we subtract the z-table value from 1)g. \(P(z \geq -3.38)\) = 0.9996h. \(P(z < 4.98)\) = essentially 1 or 100%, as the Z-table typically does not go past 3.49, and anything further is so close to 0 or 1 that it is typically recorded as such.