The corresponding z-score was found earlier, which is about 1.285. The top 10% corresponds to the 90th percentile. What is the lowest score you can earn and still be eligible to be hired by the agency? The mean is µ = 50 and standard deviation is σ = 10. An agency will only hire applicants with scores in the top 10%. Scores for the California Police Officer Standards and Training test are normally distributed, with a mean of 50 and a standard deviation of 10. Given z, to find x, use the formula x = µ + zσ. Transforming a z-score into a data value Given a z-score, it can be converted back into a data value by solving for x in the equation z= Convert 30% into a probability: 0.3000, and find the z-score corresponding to this area: between -0.53 and -0.52. The z-score that corresponds to the 30th percentile (P30 ) of the distribution’s area. Convert 84.13% into a probability: 0.8413, and find the z-score corresponding to area 1 − 0.8413 = 0.1587. The z-score that corresponds to 84.13% of the distribution’s area to its right. Find the z-score corresponding to area 1 − 0.4090 = 0.5910. The z-score that corresponds to 0.4090 of the distribution’s area to its right. The z-score that corresponds to a cumulative area of 0.8888. (You may use either of these or average them). The desired z-score is between 1.28 and 1.29. Even though this exact area is not in the table, pick the closest areas. Convert 90% into a probability (area) first: 0.9000. The z-score that corresponds to the 90th percentile (P90 ) of the distribution’s area. The z-score needed corresponds to a left area of 1 − 0.9616 = 0.0384. First convert 96.16% into a probability (area): 0.9616. The z-score that corresponds to 96.16% of the distribution’s area to its right. Thus, the z-score needed corresponds to a left area of 1 − 0.1075 = 0.8925. The table lists the cumulative area: to the left of the z-score. The z-score that corresponds to 0.1075 of the distribution’s area to its right. Look for the given area in the table and find the corresponding z-score: -0.35 b. The z-score that corresponds to a cumulative area of 0.3632 (the cumulative area is the area to the left of the z-score). Draw a picture and include a short explanation a. To make this easier, first draw a picture. The same table will be used, but you will search the center of the table to find the probability first, and then determine the z-score that corresponds to that probability. Starting with a probability, you will find a corresponding z-score. Thus, P (X > 39) ≈ 1 − 0.3085 = 0.6915ĥ.3 Normal Distributions: Finding Values Now the process from 5.2 will be reversed. We need the area to the right of the z-score. For P (X > 54), the area to the right is needed. The mean is µ = 45 and standard deviation is σ = 12. OR, using complements and the answer to part a, P (X ≤ 70) = 1 − P (X > 70) ≈ 1 − 0.1949 ≈ 0.8051 (b) What is the probability that a randomly selected vehicle is traveling under 50 miles per hour? We are interested in P (X 54). We are interested in P (X ≤ 70), thus the area to the left of this z-score can be read directly off the table: 0.8051. Since we are interested in X > 70, we need the area to the right of the z-score, thus P (X > 70) ≈ 1 − 0.8051 ≈ 0.1949 (a) What is the probability that a randomly selected vehicle is not violating the speed limit? The z-score is the same: 0.86. The area corresponding to a z-score of 0.86 in the table is 0.8051. Step 2: Find the appropriate area between the normal curve and the axis using the table: The table contains cumulative areas (to the left of the z-value). For the probability that X a, convert a into a z-score using a−µ σ and use the table to find the area to the right of the z-score.The mean (expected value) µ and standard deviation σ should be given in the problem. Your answers may vary slightly.ĥ.2 Normal Distributions: Finding Probabilities If you are given that a random variable X has a normal distribution, finding probabilities corresponds to finding the area between the standard normal curve and the x-axis, using the table of z-scores. Chapter 5: Normal Probability Distributions - Solutions Note: All areas and z-scores are approximate.