Data from the article The Osteological Paradox Problems infe

Data from the article \"The Osteological Paradox: Problems inferring Prehistoric Health from Skeletal Samples\" (Current Anthropology (1992):343-370) suggests that a reasonable model for the distribution of heights of 5-year old children (in centimeters) is N(100, 6²) . Let the letter X represent the variable \"height of 5-year old\", and use this information to answer the following. Use 4 decimal places unless otherwise indicated.

(a) P(X > 89.2) =  

(b) P(X < 109.78) =  

(c) P(97 < X < 106) =  

(d) P(X < 85.6 or X > 111.4) =  

(e) P(X > 103) =  

(f) P(X < 98.2) =  

(g) P(100 < X < 124)=  

(h) The middle 80% of all heights of 5 year old children fall between __ and __ . (Use 2 decimal places.)

Solution

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    89.2      
u = mean =    100      
          
s = standard deviation =    7.874007874      
          
Thus,          
          
z = (x - u) / s =    -1.371601372      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.371601372   ) =    0.914906217 [answer]

b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    109.78      
u = mean =    100      
          
s = standard deviation =    7.874007874      
          
Thus,          
          
z = (x - u) / s =    1.242061242      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   1.242061242   ) =    0.892893016 [answer]

c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    97      
x2 = upper bound =    106      
u = mean =    100      
          
s = standard deviation =    7.874007874      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.381000381      
z2 = upper z score = (x2 - u) / s =    0.762000762      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.351601483      
P(z < z2) =    0.776970225      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.425368742   [answer]

d)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    85.6      
x2 = upper bound =    111.4      
u = mean =    100      
          
s = standard deviation =    7.874007874      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.828801829      
z2 = upper z score = (x2 - u) / s =    1.447801448      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.03371465      
P(z < z2) =    0.926163706      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.892449056      

Thus, those outside this interval is the complement =   0.107550944   [answer]

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