For Questions 1-3 Using The Standard Normal Distribution Tab

For Questions 1-3 using The Standard Normal Distribution Table Comput

For questions 1-3: Using the standard normal distribution table, compute the following probabilities: P( Z ≥ 1.8) is: .0359 .0062 .4938 .4641

Question 2 0.5 pts P(Z ≤ 1.5) is: .8750 .9599 .9332 .9938

Question 3 0.5 pts P(1.34 ≤ Z ≤ 2.72) is: .0868 .0476 .8746 .0309

Paper For Above instruction

Probability calculations involving the standard normal distribution are fundamental in statistics, especially when dealing with data that approximately follow a normal distribution. The standard normal distribution table, also known as the Z-table, provides the probability that a standard normal variable Z is less than or equal to a particular value. This paper explores the process of computing such probabilities using the Z-table, with specific examples from the provided questions.

Question 1 involves computing P(Z ≥ 1.8). Since the Z-table provides probabilities for P(Z ≤ z), we start by looking up P(Z ≤ 1.8). From the standard normal distribution table, P(Z ≤ 1.8) is approximately 0.9641. Because the total probability is 1, the probability that Z is greater than or equal to 1.8 is: P(Z ≥ 1.8) = 1 - P(Z ≤ 1.8) = 1 - 0.9641 = 0.0359. This means there is about a 3.59% chance that Z exceeds 1.8. The options given align with this calculation, making 0.0359 the correct choice.

Question 2 asks for P(Z ≤ 1.5). Referencing the Z-table, P(Z ≤ 1.5) is approximately 0.9332. Among the options, 0.9332 directly corresponds to this value, indicating that there's a 93.32% probability that Z is less than or equal to 1.5. Such calculations are essential in assessing likelihoods for normally distributed variables, especially when establishing confidence intervals or hypothesis testing.

Question 3 is more involved, asking for the probability that Z lies between 1.34 and 2.72: P(1.34 ≤ Z ≤ 2.72). To compute this, we find P(Z ≤ 2.72) and P(Z ≤ 1.34) from the Z-table. P(Z ≤ 2.72) is approximately 0.9967, and P(Z ≤ 1.34) is approximately 0.9099. The probability that Z falls within this interval is the difference: 0.9967 - 0.9099 = 0.0868. This indicates about an 8.68% chance for a Z-score in this range. The closest matching option is 0.0868.

In summary, proficiency with the standard normal distribution table enables precise computation of probabilities related to normally distributed variables. Such computations are crucial in diverse statistical applications, including determining probabilities, confidence levels, and critical regions for hypothesis testing. The approach involves identifying the relevant Z-scores, consulting the table for cumulative probabilities, and calculating differences where necessary. Mastery of this process enhances one's ability to interpret and analyze data within the framework of standard normal assumptions.

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