Using the table for finding the areas under normal curves fi

Using the table for finding the areas under normal curves, find the area under a noraml curve with a mean of 200 and a standard deviation of 10 between the values of:

a. 200 to 205

b. 195 to 205

c. 200 to 215

d. 195 to 215

e. 186.5 to 217

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    200      
x2 = upper bound =    205      
u = mean =    200      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.191462461 [answer]

b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    195      
x2 = upper bound =    205      
u = mean =    200      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.5      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.308537539      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.382924923   [answer]

c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    200      
x2 = upper bound =    215      
u = mean =    200      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.433192799   [answer]

d)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    195      
x2 = upper bound =    215      
u = mean =    200      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.5      
z2 = upper z score = (x2 - u) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.308537539      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.62465526   [answer]

e)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    186.5      
x2 = upper bound =    217      
u = mean =    200      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.35      
z2 = upper z score = (x2 - u) / s =    1.7      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.088507991      
P(z < z2) =    0.955434537      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866926546   [answer]