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Z-score Standard Normal Distribution 0.4 -3σ -2σ σ σ 2σ 3σ Mean = 0 A Standard Normal Distribution is a normal distribution which has mean = 0 and σ = 1. (See statisticsTheory.doc) Z - Scores The Z-Score enable transformation of any normal distribution to standard normal distribution by using a formula: Z= x-μ σ Formula (1) The above formula is for obtaining a z-score for an entire population. Usability testing obviously samples a very small subset of the population and thus the following formula is used: Z= x-x σ Formula (2) and s are used as estimators for the population's true mean and standard deviation. Both formulas essentially calculate the same thing: Z= data _ po int - mean s tan dard _ deviation Formula (3) © Copyrigh (2007)William Mak – All rights are reserved 1 Z-score How to Use the Z-score for Finding Probabilities The best way is to use examples to show. Consider a production process makes shafts which are normally distributed with an average diameter of 2.0cm and a variance of 4 cm sq.. Find the probabilities that a single bolt selected at random from the process will have a diameter of: (a) (b) (c) (d) more than 3 cm; exactly 2 cm; less than 1 cm; Between 1.5 and 3cm. Solution: Let x = diameter of a shaft ~ N (2, 4), where N (2, 4) means that the random variable is normal distribution with mean value = 2, and variance = 4. Remember that sample mean, x = And (a) s 2 ∑x n ∑x = n 2 - x 2 For P (x > 3): Area = 0.1915 By Formula (1): Z= x-μ = 0.5 σ Area = 0.3085 Therefore, P(x > 3) = P (Z > 0.5) Z = 0.5 Now, using Z = 0.5, we can find from Standard Normal Distribution Table that the area between 0 and 0.5 (as the diagram above) is 0.1915. So, the area enclosed beyond Z = 0.5 is equal to 0.3085. Therefore the probability that the diameter of a shaft is bigger than 3 cm is 0.3085 or 31%. Remember that the total area enclosed by the curve is 1, or 100%. (b) diameter is exactly = 2 cm: © Copyrigh (2007)William Mak – All rights are reserved 2 Z-score P (x = 2) Area = 0.000 By Formula (1): Z= 2-2 =0 2 Z=0 Therefore, P (x = 2) = P (Z = 0) For Z = 0.5, using Standard Normal Distribution Table the area between Z = 0 and Mean (which is 0), as the diagram above, is 0.000. Hence, P (x = 2) = P (Z = 0) = 0 (c) P (1.5 < x < 3) =? When x = 1.5, using the transformation formula, Z = Similarly, Z = 1.5 - 2 = -0.25 2 3-2 = 0.5 2 Therefore, P (1.5 < x < 3) = P (-0.25 < Z < 0.5) = P (-0.25< Z < 0) plus P (0 < Z < 0.5) the area = 0.987 + 0.1915 = 0.2902 So, the probability = 29% -0.25 0.5 © Copyrigh (2007)William Mak – All rights are reserved 3 Z-score Example again! (2) For the process in Example 1, if 16 shafts are selected what is the probability that the average diameter will exceed 2.1 cm? Solution: Let x bet the diameter of a shaft from a population of shafts ~ N (μ, σ ) 2 Let σ x = average diameter, of 16 bolts ~ N (μ, 2 n ) 2 ~ N (μ, 2 16 1 ~ N (μ, ) 4 Note: σ ) 2 is the variance of the 16 pieces of shafts having the mean = x . Hence n the standard deviation of that 16 pieces of shaft = σ n σ n 2 = σ n . is termed as standard error For P ( x > 2.1) Z= data_point - μ s tan dard _ deviation Z= x-μ σ x 2.0 2.1 n 2.1 - 2 2.1 - 2 1 4 = 0.2 = 0 Z ? So, P ( x > 2.1) = P (Z > 0.2) From table, the area being = 0.5 – 0.0793 = 42% © Copyrigh (2007)William Mak – All rights are reserved 4 Z-score Theorem (Central Limit) If x , x ,x 1 2 3 … x 4 are randomly selected and each of which has a mean, μ, and variance, σ , then: 2 (a) x = ( (b) x + x +x 1 2 ∑x = ( x1 + 3 … x +x 2 3 x )/n= n … ∑x n σ ---- N (μ, n 2 ) x ) --- N (n μ, n σ ) 2 n Example (3) The average pay of employees in a company is $100 per week, having standard deviation of $4 per week. Find the probabilities that: (a) the average salary of 9 employees is less than $98.5 per week (b) the total salary bill of 25 employees is less than $2600 per week. Solution: σ (a) Let x = average salary of 9 employees ~ N (100, n 2 ) 2 ~ N (100, 4 9 ) For P ( x < 98.5) data_point - μ Z= s tan dard _ deviation = = x-μ S .D 98.5 - 100 2 4 9 = - 1.125 0 -1.125 © Copyrigh (2007)William Mak – All rights are reserved Z 5 Z-score ∴ P ( x < 98.5) = P (Z < -1.125) = P (Z > 1.125) = 0.5 – 0.3697 (as from table) = 13% (b) Let ∑x be the total salary bill of 25 employees ~ (n μ, n σ ) 2 2 ~ (25x100, 16x 4 ) For P ( ∑x < 2600) Z= data_point - μ s tan dard _ deviation Area =? = 2600 - 25 x100 2 =1 25 x 4 ∴ P ( ∑x < 2600) = P (Z < 1) 0 1 Z Area = 0.3413 + 0.5 = 0.8413 = 84.13% © Copyrigh (2007)William Mak – All rights are reserved 6