Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 28.4 kilograms and standard deviation ? = 4.1 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x intervals to z intervals. (Round your answers to two decimal places.) x < 30 z < ____   19 < x ___ < z     32 < x < 35 ____ < z < ____

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Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 28.4 kilograms and standard deviation ? = 4.1 kilograms. Let x be the weight of a fawn in kilograms.

Convert the following x intervals to z intervals. (Round your answers to two decimal places.)

x < 30
z < ____
 
19 < x
___ < z
 
  32 < x < 35
____ < z < ____


Convert the following z intervals to x intervals. (Round your answers to one decimal place.)

 −2.17 < z
___ < x

z < 1.28
x < ____

 −1.99 < z < 1.44
 ____< x < ____
 
 If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and the figure above.
Yes. This weight is 3.51 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
Yes. This weight is 1.76 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.    
No. This weight is 3.51 standard deviations below the mean; 14 kg is a normal weight for a fawn.
No. This weight is 3.51 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
No. This weight is 1.76 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
 
 If a fawn is unusually large, would you say that the z value for the weight of the fawn will be close to 0, −2, or 3? Explain.
It would have a negative z, such as −2.
It would have a large positive z, such as 3.   
 It would have a z of 0.
The Standard Normal Distribution
(u = 0, o = 1)
%3D
3
-2
3
68% of area
95% of area
99.7% of area
Transcribed Image Text:The Standard Normal Distribution (u = 0, o = 1) %3D 3 -2 3 68% of area 95% of area 99.7% of area
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