If X is normally distributed with a mean of 6 and a standard

If X is normally distributed with a mean of 6 and a standard deviation of 2. Find the followings probabilities.

Solution

1.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    2      
x2 = upper bound =    10      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736   [ANSWER]

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2.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    8      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    1      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1   ) =    0.158655254 [ANSWER]

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3.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    -3      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -3   ) =    0.001349898 [ANSWER]

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4.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    3      
x2 = upper bound =    9      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.5      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597   [ANSWER]

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5.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    7      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    0.5      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.5   ) =    0.308537539 [ANSWER]

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6.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1      
u = mean =    6      
          
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) / s =    -2.5      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.5   ) =    0.006209665 [ANSWER]