Heights of 20yearold men vary approximately according to a N

Heights of 20-year-old men vary approximately according to a Normal distribution, such that X ~ N(=176.9 cm, = 7.1 cm).

Determine the range of scores that fall within 1 standard deviation of the mean. Round your answer to the nearest tenth.

What percentage of scores fall within this range? Convert a height of 165.9 cm to a z-score. Round your answer to the nearest hundredth.

Solution

Determine the range of scores that fall within 1 standard deviation of the mean. Round your answer to the nearest tenth.

Adding and subtracting the standard deviation to the mean,

u - sigma = 176.9 - 7.1 = 169.8

u + sigma = 176.9 + 7.1 = 184

Thus, it is between (169.8, 184) [ANSWER]

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What percentage of scores fall within this range?
We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    169.8      
x2 = upper bound =    184      
u = mean =    176.9      
          
s = standard deviation =    7.1      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492 = 68.27% [ANSWER]

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Convert a height of 165.9 cm to a z-score. Round your answer to the nearest hundredth.

As

z = (x-u)/sigma

Then

z = (165.9 - 176.9)/7.1 = -1.549295775 = -1.55 [ANSWER]