Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 29.4 kilograms and standard deviation σ = 4.2 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
-0.65  < 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.

Yes. This weight is 3.67 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.Yes. This weight is 1.83 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.    No. This weight is 3.67 standard deviations below the mean; 14 kg is a normal weight for a fawn.No. This weight is 3.67 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.No. This weight is 1.83 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.

It would have a large positive z, such as 3.It would have a negative z, such as −2.    It would have a z of 0.

Answer 1

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 29.4 kilograms and standard deviation σ = 4.2 kilograms. Let x be the weight of a fawn in kilograms.

We know that will have a standard Normal distribution.

(a)    x < 30.

When x=30, we have
z < 0.14

(b)    19 < x

When X=19, we have
-2.48 < z

(c)    32 < x < 35

When X=32, we have

When X=35, we have
32<x<35= 0.62 < z <1.33  

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

(d)    −2.17 < z

From we have

When Z=-2.17, X=29.4-2.17*4.2=29.4-9.114=20.286

20.29<X

(e)    z < 1.28

When Z=1.28, X=29.4+1.28*4,2=29.4+5.376=34.776
x <34.78

(f)    −1.99 < z < 1.44

When Z=-1.99, X=29.4-1.99*4.2=29.4-8.358=21.042

When Z=1.44,X=29.4+1.44*4.2=29.4+6.048=35.448
21.04< x <35.45

g) 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 3.67 standard deviations below the mean; 14 kg is an unusually low 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.

It would have a large positive z, such as 3.