Suppose that a random variable x is normally distributed wit

Suppose that a random variable x is normally distributed with mean 50 and standard deviation 9.

(give answer a to four decimal places)

(give answer b to two decimal places)

(give answer c to four decimal places)

Please make the final answer in bold so I can easily find it

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    44      
x2 = upper bound =    67      
u = mean =    50      
          
s = standard deviation =    9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.666666667      
z2 = upper z score = (x2 - u) / s =    1.888888889      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.252492538      
P(z < z2) =    0.970546641      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.718054103   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.8888      
          
Then, using table or technology,          
          
z =    1.220171153      
          
As x = u + z * s,          
          
where          
          
u = mean =    50      
z = the critical z score =    1.220171153      
s = standard deviation =    9      
          
Then          
          
x = critical value =    60.98154037   [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    50.2      
u = mean =    50      
n = sample size =    25      
s = standard deviation =    9      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.111111111      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.111111111   ) =    0.455764119 [ANSWER]