Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 27.2 kilograms and standard deviation σ = 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.)

(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.

Yes. This weight is 3.22 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.Yes. This weight is 1.61 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.     No. This weight is 3.22 standard deviations below the mean; 14 kg is a normal weight for a fawn.No. This weight is 3.22 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.No. This weight is 1.61 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 z of 0.It would have a large positive z, such as 3.     It would have a negative z, such as −2.

Answer 1

Given that the Fawns between 1 and 5 months old have a bodyweight that is approximately normally distributed with mean μ = 27.2 kilograms and standard deviation σ = 4.1 kilograms.

Now the Z score is calculated as:

(a)    x < 30 at X = 30

z < 0.68

(b)    19 < x at X = 119

-20 < z

(c)    32 < x < 35 at X = 32 and 35

Thus converted into Z scores as:

1.19< z <1.90

Converting the following z intervals to x intervals.

(d)    −2.17 < z

18.3< x

(e)    z < 1.28

x < 37.4

(f)    −1.99 < z < 1.44

Hence 19.0 < x < 37.1

(g) Given that a fawn weighs 14 kilograms, so, to make it an unusually small animal we need to find the Z score at X = 14 and if the Z score is below -2 or above 2 then it is called as an unusually small.

The Z score is calculated as:

Thus, the conclusion is:

Yes. This weight is 3.22 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

(h) If a fawn is unusually large, we would say that the z value for the weight of the fawn will be close to 3, because the Z score at 3 or larger is very far away from the mean of the distribution and its probability is negligible.

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

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