Assume the average weight of an adr is normally distributed with a mean of 69.5

Question

Assume the average weight of an adr is normally distributed with a mean of 69.5 and standard deviation of 2.4. A sample of64 males is taken from population and their height is measured. What is the prob the sample average falls between 69.2 and 69.95? Assume the average weight of an adr is normally distributed with a mean of 69.5 and standard deviation of 2.4. A sample of64 males is taken from population and their height is measured. What is the prob the sample average falls between 69.2 and 69.95?

Explanation / Answer

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    69.2      
x2 = upper bound =    69.95      
u = mean =    69.5      
n = sample size =    64      
s = standard deviation =    2.4      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.774537545   [ANSWER]  

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