Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 29.4 kilograms and standard deviation ? = 4.4 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.) (a)    x < 30 z < ___ (b)    19 < x  ______< z (c)    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 ? = 29.4 kilograms and standard deviation ? = 4.4 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.)

(a)    x < 30
z < ___

(b)    19 < x
 ______< z

(c)    32 < x < 35
 ____< z < ____


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

(d)    −2.17 < z
_____ < x

(e)    z < 1.28
x < ______

(f)    −1.99 < z < 1.44
 ______< x < ______
(g) If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and the figure above.
A-Yes. This weight is 3.50 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
B-Yes. This weight is 1.75 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.   
C-No. This weight is 3.50 standard deviations below the mean; 14 kg is a normal weight for a fawn.
D-No. This weight is 3.50 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.No. This weight is 1.75 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
(h) 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.
A-It would have a z of 0.
B-It would have a large positive z, such as 3.    
C-It would have a negative z, such as −2.
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