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area. Step 4: State the conclusion. :The area between z = -2.84 and z = 0.95. is 0.8266 or 82.66% . III. COMPUTING PROBABILITIES AND PERCENTILES USING THE STANDARD NORMAL TABLE Probabilities and percentiles can be solved using the standard normal table. The standard score is a measure of relative standing. It is a standard deviation value. It is the distance between the given value X and the mean. Any normal distribution can be transformed to a standard normal distribution and vice versa by the formula. Each of the values obtained using the given formula is called z-scores. Unknown Value Population Sample x- # z-score Z = - x- x Score x =zo+ # x - zs + x Mear 1= X -20 X=X- ZS Standard Deviation x- H x- x 0 =- S = - Illustrative Example 4: The number of mangrove trees planted by the volunteers is 679 per year with a standard deviation of 76. What is the probability that the volunteers will plant more than 850 trees per year? Step 1: Write the given and x = 850 # = 679 unknown. 0 = 76 Z =? Step 2: Apply the formula. x- H 850- 679 Z =_ -= 2.25 76 Step 3: Locate the area of the z- P(x > 850) = P(Z > 2.25) value from the z-table = 1.0000 - 0.9878 = 0.0122 or 1.22% Step 4: Sketch the normal curve 1.22% and shade the desired area Step 5: State the conclusion : The probability that the volunteers will plant more than 850 trees per year is 1.22% Illustrative Example 5: The sample average length of a sock that is produced at a factory is 34 cm and the sample standard deviation is 6. What is the probability that the length of a sock will be below 22 cm? 3 GSC-CID-LRMS-ESSLM, v.r. 03.00, Effective June 14, 2021 Step 1: Write the given and x = 22 X = 34 unknown. S =6 Z =? Step 2: Apply the formula. Z - K-X 22-34 -12-2 Step 3: Locate the area of the z- P(2