Use the normal distribution of IQ scores, which has a mean of 95 and a standard

Question

Use the normal distribution of IQ scores, which has a mean of 95 and a standard deviation of 17, and the following table with the standard scores and percentiles for a normal distribution to find the indicated quantity.

Percentage of scores greater than 52.5 is ____%.

(Round to two decimal places as needed.)

99.98

Standard score Percent -3.0 0.13 -2.5 0.62 -2 2.28 -1.5 6.68 -1 15.87 -0.9 18.41 -0.5 30.85 -0.1 46.02 0 50.00 0.10 53.98 0.5 69.15 0.9 81.59 1 84.13 1.5 93.32 2 97.72 2.5 99.38 3 99.87 3.5

99.98

Explanation / Answer

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    52.5      
u = mean =    95      
          
s = standard deviation =    17      
          
Thus,          
          
z = (x - u) / s =    -2.5      
          
Thus, using the table, the right tailed area of this is          
          
P(z >   -2.5   ) =    100 - 0.62 = 99.38% [ANSWER]

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