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

The probabilities for \(P(z<2.36)\) and \(P(z \leq 2.36)\) are 0.9910. The probability for \(P(z<-1.23)\) is 0.1093. The probability for \(1.142)\) is 0.0228. The probability for \(P(z \geq -3.38)\) is 0.0003. The probability for \(P(z<4.98)\) is approximately 1.

Step-by-step solution

Determining the probabilities

Firstly, recall that the z-score of a standard normal distribution N(0,1) is already given. To determine the probabilities, use the z-table: these are normally given, showing the probabilities of each z-score. The z-scores have already been found for each of these probabilities.

Calculate probability for \(P(z

Locate 2.36 inside the table. Follow that row until you hit the column labeled '.00'. The intersection is the probability answer, 0.9910.

Calculate probability for \(P(z \leq 2.36)\)

This probability is the same as \(P(z<2.36)\) because the distribution is continuous, thus including the endpoint doesn't change the probability. Hence, the probability is also 0.9910.

Calculate probability for \(P(z

Because the standard normal distribution is symmetric around 0, \(P(z<-1.23)\) will correspond to \(1-P(z<1.23)\). So, find the probability of 1.23 in the z-table, which is 0.8907. Therefore, the probability of \(P(z<-1.23)\) is \(1-0.8907 = 0.1093\).

Calculate probability for \(P(1.14

To find this probability, find the probabilities of 3.35 and 1.14 in the z-table, which are 0.9996 and 0.8729 respectively. Subtract these two probabilities to find the probability between these two z-scores: \(0.9996 - 0.8729 = 0.1267\).

Calculate probability for \(P(-0.77 \leq z \leq -0.55)\)

Find the probabilities of \(-0.55\) and \(-0.77\) in the z-table, which are 0.7054 and 0.2206 respectively. Subtract these two probabilities to find the required probability: \(0.7054 - 0.2206 = 0.4848\).

Calculate probability for \(P(z>2)\)

This corresponds to \(1-P(z<2)\). So, find the probability of 2 in the z-table, which is 0.9772. Therefore, the probability of \(P(z>2)\) is \(1-0.9772 = 0.0228\).

Calculate probability for \(P(z \geq -3.38)\)

This corresponds to \(1-P(z<-3.38)\) because of the symmetry of the standard normal distribution. Find the probability of -(-3.38) in the z-table, which is 0.9997. Therefore, the probability of \(P(z \geq -3.38)\) is \(1-0.9997 = 0.0003\).

Calculate probability for \(P(z

For values above 3 in the z-table, the probabilities are essentially 1, so we can assume that \(P(z < 4.98)\) is approximately 1.