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In the United States, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. If height is normally distributed, which of the following percentages represents the number of women in this country who are between 63 and 70.5 inches tall? A. 82%

Respuesta :

princessalane
Easiest way to find probability of women between 63 and 70.5 inches tall is to find probability < 70.5 and subtract the probability < 63

 

Easiest way to find probability of of 63 and 70.5 is to calculate z score of each

 

z score of 63 = (x-mean)/std = (63-65.5)/2.5 = -1 using a z table gives us prob < 63 is .1587

 

z score of 70.5 = (x-mean)/std = (70.5-65.5)/2.5 = 2 using a z table gives us prob < 70.5 is .9772

 

Subtract the .1587 from .9772 gives us .8185 or 81.85%

lhmarianateixeira

In this exercise we have to use the knowledge of probability, so we find that:

[tex]81.85\%[/tex]

Given the text information we can calculate the probability as;

  • probability of women between 63 and 70.5 inches tall is to find probability < 70.5 and subtract the probability < 63

[tex]z\ score \ of \ 63 = (x-mean)/std = (63-65.5)/2.5 = -1\\ z < 63 =1587\\ z \ score \ of \ 70.5 = (x-mean)/std = (70.5-65.5)/2.5 = 2\\ z< 70.5 =0.9772 [/tex]

See more about probability at brainly.com/question/795909

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