Assume that womens heights arc normally distributed with a m

Assume that women\'s heights arc normally distributed with a mean given by Mu - 63.8 in, and a standard deviation given by sigma = 2.6 in. (a) If 1 woman is randomly selected, find the probability that her height is between 63.4 in and 64.4 in The probability is approximately [ ] . (Round to tour decimal places as needed.) (b) If 11 women an randomly selected, find the probability that they have a mean height between 63.4 a sad 64.4 in. The probability is approximately [ ] . (Round to four decimal plates as needed.) (c) Why can the central limit theorem be used in part (b), even though the maple site does not exceed 30? A The sample site needs to be less than 30, not greater than 30. B. The population size is greater than 30. D. The sample is normally distributed. E The population a normally distributed.

Solution

a)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    63.4      
x2 = upper bound =    64.4      
u = mean =    63.8      
          
s = standard deviation =    2.6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.153846154      
z2 = upper z score = (x2 - u) / s =    0.230769231      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.438865521      
P(z < z2) =    0.591252957      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.152387436   [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    63.4      
x2 = upper bound =    64.4      
u = mean =    63.8      
n = sample size =    11      
s = standard deviation =    2.6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.510249968      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.765374952      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.304938175      
P(z < z2) =    0.777975848      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.473037673   [ANSWER]

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c)

OPTION D: The population is normally distributed.