A sample of concrete specimens of a certain type is selected, and the compressiv

Question

A sample of concrete specimens of a certain type is selected, and the compressive strength of each specimen is determined. The mean and standard deviation are calculated as x = 600C and s = 500, and the sample histogram is found to be well approximated by a normal curve. Approximately what percentage of the sample observations are between 5500 and 6500? (Round the answer to the nearest whole number.) Approximately what percentage of sample observations are outside the interval from 5000 to 7000? (Round the answer to the nearest whole number.) What can be said about the approximate percentage of observations between 5000 and 5500? (Round the answer to the nearest whole number.) Why would you not use Chebyshev's Rule to answer the questions posed in parts (a)-(c)?_ Chebyshev's Rule does not apply to the normal distribution. Chebyshev's Rule is the best way to solve this problem. Chebyshev's Rule does not apply to values in these ranges. Chebyshev's Rule is not used because the histogram is well approximated by a normal curve. Chebyshev's Rule typically gives much larger values than are appropriate for a normal distribution.

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5500      
x2 = upper bound =    6500      
u = mean =    6000      
          
s = standard deviation =    500      
          
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% [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5000      
x2 = upper bound =    7000      
u = mean =    6000      
          
s = standard deviation =    500      
          
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 = 95% [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    5000      
x2 = upper bound =    5500      
u = mean =    6000      
          
s = standard deviation =    500      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    -1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.158655254      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.135905122 = 14% [ANSWER]

***********************

d)

OPTION D: Chebyshev's Rule is not used because the histogram is well approximated by a normal curve. [ANSWER]

We only use Chebyshev's Rule to get the least percentage, without knowing the distirbution involved. Since we know it is a normal distirbution, we don't use Chebyshev's Rule.
      
      

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