Fawns between 1 and 5 months old have a body weight that is

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean = 27.1 kilograms and standard deviation = 3.1 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.)

(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 4.23 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

B. Yes. This weight is 2.11 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.  

C. No. This weight is 4.23 standard deviations below the mean; 14 kg is a normal weight for a fawn.

D. No. This weight is 4.23 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.

E. No. This weight is 2.11 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.

Solution

a)

z = (x-u)/sigma

Then if x = 30,

z = (30-27.1)/3.1 = 0.94

Thus,

z < 0.94 [ANSWER]

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b)

z = (x-u)/sigma

Then if x = 19,

z = (19-27.1)/3.1 = -2.61

Thus,

-2.61 < z [ANSWER]

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c)

z = (x-u)/sigma

Then if x = 32,

z = (32-27.1)/3.1 = 1.58

If x = 35,

z = (35-27.1)/3.1 = 2.55

Thus,

1.58 < z < 2.55 [ANSWER]

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d)

As

x = u + z*sigma

Then if z = -2.17,

x = 27.1 + (-2.17)*3.1 = 20.4

Thus,

20.4 < x [ANSWER]

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