Hi Im having quite a bit of difficulty with this question It

Hi, Im having quite a bit of difficulty with this question. It requires some approach using R aswell. Any help will be gladly appreciated.

Solution

a)

The convention is to write N(mean, variance).

Hence, here,

X~N(220, 12^2) mm or

X~N(220, 144) mm [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    192      
u = mean =    220      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    -2.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -2.333333333   ) =    0.009815329 [ANSWER]

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C)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    180      
x2 = upper bound =    195      
u = mean =    220      
          
s = standard deviation =    12      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3.333333333      
z2 = upper z score = (x2 - u) / s =    -2.083333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.00042906      
P(z < z2) =    0.018610425      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.018181365 [ANSWER]

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d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.0475      
          
Then, using table or technology,          
          
z =    -1.669592577      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    220      
z = the critical z score =    -1.669592577      
s = standard deviation =    12      
          
Then          
          
x = critical value =    199.9648891   [ANSWER]

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