Some IQ tests are standardized to a Normal distribution with

Some IQ tests are standardized to a Normal distribution with mean 100 and standard deviation 16.

(a)What proportion of IQ scores is between 95 and 105?

(b)What is the 80th percentile of the IQ scores?

(c)A random sample of 10 candidates are about to take the test. What is the probability that at least half of them will score between 95 and 105?

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    95      
x2 = upper bound =    105      
u = mean =    100      
          
s = standard deviation =    16      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.3125      
z2 = upper z score = (x2 - u) / s =    0.3125      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.377330282      
P(z < z2) =    0.622669718      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.245339437   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8      
          
Then, using table or technology,          
          
z =    0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    100      
z = the critical z score =    0.841621234      
s = standard deviation =    16      
          
Then          
          
x = critical value =    113.4659397   [ANSWER]

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c)

This means at least 5 of them score 95 to 105.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    10      
p = the probability of a success =    0.245339437      
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.927198461
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.072801539 [ANSWER]