The scores for a certain test of intelligence are normally distributed with mean
ID: 2932562 • Letter: T
Question
The scores for a certain test of intelligence are normally distributed with mean 125 and standard deviation 16. Find the 90th percentile Click the icon to view the table of standard scores and percentiles. The 90th percentile is (Round to the nearest whole number as needed)
Standard Scores and Percentiles z-score Percentile z-score Percentile z-score Percentile z-score Percentile 50.00 00.02 00.13 00.19 1.00 15.87 0.00 86.43 0.95 17.11 0.05 51.99 12 88.49 90.32 91.92 93.32 -2.6 00.47 0.75 2266 0.25 59.87 16 94.52 95.54 96.41 97.13 97.72 98.21 -2.0 02.28 0.45 32.64 0.55 70.88 2.298.61 98.93 99.18 99.38 99.53 99.65 99.74 99.81 99.87 99.98 -3.0 2.9 0.90 18.41 -2.8 00.26 0.85 19.77 0.15 55.96 0.80 21.19 53.98 2.7 00.35 0.20 57.93 2.5 00.62 0.30 61.79 0.70 24.20 -2.4 00.82 0.65 25.78 0.35 63.68 0.60 27.43 2.3 01.07 0.40 65.54 -2.2 01.39 0.55 29.12 0.45 67.36 2.0 30.85 2.1 01.79 0.50 0.50 69.15 72.57 1.8 03.59 -0.35 36.32 0.65 74.22 75.80 1.6 05.48 -0.25 40.13 0.75 77.34 78.81 1.4 08.08 0.15 44.04 0.85 80.23 81.59 0.05 48.01 0.95 82.89 84.13 1.9 02.87 0.40 34.46 1.7 04.46 0.30 38.21 1.5 06.68 0.20 42.07 0.90 09.68 11.51 13.57 46.02 1.3 1.2 0.10 0.00 50.00 1.00Explanation / Answer
Mean is 125 and s is 16
For 90th percentile, from the above table, the z value is 1.3
thus answer is mean+s*z=125+16*1.3=145.8