EXAMPLE 1.45

How high for the top 10%? Scores for college-bound students on the SAT Critical Reading test in recent years follow approximately the N(500, 120) distribution.33 How high must a student score to place in the top 10% of all students taking the SAT?

Again, the key to the problem is to draw a picture. Figure 1.29 shows that we want the score x with an area of 0.10 above it. That’s the same as area below x equal to 0.90.

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Figure 1.29 Locating the point on a Normal curve with area 0.10 to its right, Example 1.45.

Statistical software has a function that will give you the x for any cumulative proportion you specify. The function often has a name such as “inverse cumulative probability.” Plug in mean 500, standard deviation 120, and cumulative proportion 0.9. The software tells you that x = 653.786. We see that a student must score at least 654 to place in the highest 10%.

Without software, first find the standard score z with cumulative proportion 0.9, then “unstandardize” to find x. Here is the two-step process:

  1. 1. Use the table. Look in the body of Table A for the entry closest to 0.9. It is 0.8997. This is the entry corresponding to z = 1.28. So z = 1.28 is the standardized value with area 0.9 to its left.

  2. 2. Unstandardize to transform the solution from z back to the original x scale. We know that the standardized value of the unknown x is z = 1.28. So x itself satisfies

    66

    Solving this equation for x gives

    x = 500 + (1.28)(120) = 653.6

    This equation should make sense: it finds the x that lies 1.28 standard deviations above the mean on this particular Normal curve. That is the “unstandardized” meaning of z = 1.28. The general rule for unstandardizing a z-score is

    x = μ + zσ