Stanford - Binet IQ Test scores are normally distributed with a mean score of 10

Question

Stanford - Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 15. Write the equation that gives the z score corresponding to a Stanford-Binet IQ test score. Find the probability that a randomly selected person has an IQ test score. (Round your answers to 4 decimal places.) Suppose you take the Stanford-binet IQ Test and receive a score of 140. What percentage of people would receive a some higher than yours? (Round your answer to 2 decimal places.)

Explanation / Answer

b)

As z = (x-u)/sigma, then

z = (x-100)/15 [ANSWER]

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c)

1.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    141      
u = mean =    100      
          
s = standard deviation =    15      
          
Thus,          
          
z = (x - u) / s =    2.733333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.733333333   ) =    0.003134842 [ANSWER]

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2.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    84      
u = mean =    100      
          
s = standard deviation =    15      
          
Thus,          
          
z = (x - u) / s =    -1.066666667      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.066666667   ) =    0.143061192 [ANSWER]

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3.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    85      
x2 = upper bound =    115      
u = mean =    100      
          
s = standard deviation =    15      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) = 0.841344746-0.158655254 =   0.682689492   [ANSWER]

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4.

z1 = lower z score =    -3      
z2 = upper z score =     3      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.998650102      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.997300204   [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    140      
u = mean =    100      
          
s = standard deviation =    15      
          
Thus,          
          
z = (x - u) / s =    2.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.666666667   ) =    0.003830381 = 0.383% [ANSWER]
     

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