Stanford–Binet IQ Test scores are normally distributed with a mean score of 100

Question

Stanford–Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 13

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 132. What percentage of people would receive a score higher than yours? (Round your answer to 2 decimal places.)

Stanford–Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 13

Find the probability that a randomly selected person has an IQ test score. (Round your answers to 4 decimal places.)

1. P(x > 150) 2. P(x < 90) 3. P(87 < x < 113)    =    4. P(-2.10 < z < 2.10)

Suppose you take the Stanford–Binet IQ Test and receive a score of 132. What percentage of people would receive a score higher than yours? (Round your answer to 2 decimal places.)

. . .

Explanation / Answer

1.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    150      
u = mean =    100      
          
s = standard deviation =    13      
          
Thus,          
          
z = (x - u) / s =    3.846153846      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.846153846   ) =    5.99932*10^-5 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    90      
u = mean =    100      
          
s = standard deviation =    13      
          
Thus,          
          
z = (x - u) / s =    -0.769230769      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.769230769   ) =    0.220878164 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    87      
x2 = upper bound =    113      
u = mean =    100      
          
s = standard deviation =    13      
          
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) =    0.682689492   [ANSWER]
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d)

z1 = lower z score =    -2.1      
z2 = upper z score =     2.1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.017864421      
P(z < z2) =    0.982135579      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.964271159   [ANSWER]

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E)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    132      
u = mean =    100      
          
s = standard deviation =    13      
          
Thus,          
          
z = (x - u) / s =    2.461538462      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.461538462   ) =    0.0069171 = 0.69% [ANSWER]
  
  

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