Probability Distributions - Oxford University Press

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196 Business Statistics Using Excel Example 5.12 Using the data from Example 5.10, if a variable X varies as a normal distribution with a mean of 100 and a standard deviation of 5, then the value of Z when X = 110 would be equal to Z= ( 110 − 100) 5 =+ 2 . The value of P(Z ≥ 2) can be calculated using Excel’s NORMSDIST () function. Note the difference between NORMDIST and NORMSDIST. Excel solution—Example 5.12 The Excel solution is il lustrated in Figure 5.18. ➜ Excel Solution Figure 5.18 Mean = Cell C5 Value Standard deviation Cell C6 Value X = Cell C8 Value P(X 110) = Cell C12 Formula: =1–C10 Z = Cell C14 Formula: =(C8–C5)/C6 P(Z +2) = Cell C16 Formula: =1–C15 NORMDIST P(Z 2) = 0.02275 Z This solution can be represented graphically by Figure 5.19. From Excel, the NORMSDIST() function can be used to calculate P(Z ≥ +2) = 0.02275.
❉ Interpretation We observe that the probability that an individual tube length is at least 110 cms is 0.02275 or 2.3% (P(X ≥ 110) = 0.02275). Note 1. This method is used to solve problems using tables of Z values and associated probabilities. 2. The value of the Z score can be calculated using the Excel function STANDARDIZE(). 3. The Excel function NORMDIST() calculates the value of the normal distribution for the specified mean and standard deviation. 4. The Excel function NORMSDIST() calculates the value of the normal distribution for the specified Z score value. Example 5.13 If we re-consider Example 5.11 and transform the value of X to Z then we find the solution is as follows: Excel solution—Example 5.13 The Excel solution is illustrated in Figure 5.20. Figure 5.20 Note we can use the Excel function STANDARDIZE to calculate the values of Z (see chapter 6). <strong>Probability</strong> <strong>Distributions</strong> 197
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Page 7 and 8: Example 5.3 Suppose a fair die is t
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Page 37 and 38: Factorials r! Cells E23:E26 Values
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Page 61: SD( X) = VAR( X ) (5.8) 1 fX 2 exp

Probability Distributions - Oxford University Press

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