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

ID: 3181804 • Letter: F

Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean = 29.9 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 < 1

(b)    19 < x
2 < z

(c)    32 < x < 35
3 < z < 4


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

(d)    2.17 < z
5 < x

(e)    z < 1.28
x < 6

(f)    1.99 < z < 1.44
7 < x < 8
(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 3.61 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. Yes. This weight is 1.81 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.     No. This weight is 3.61 standard deviations below the mean; 14 kg is a normal weight for a fawn. No. This weight is 3.61 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. No. This weight is 1.81 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.

The Standard Normal Distribution 68% of area 95% of area 99.7% of area

Explanation / Answer

1)

mean = 29.9 , std. deviation = 4.4

(a)    x < 30

z( 30 ) = ( 30 - 29.9) / 4.4 = 0.022

z < 1

x(z) = std.deviation * z + mean

x( 1) = 4.4 * 1 + 29.9 = 34.3

(b)    19 < x

z( 19 ) = ( 19 - 29.9) / 4.4 = -2.47

2 < z

x(z) = std.deviation * z + mean

x( 2) = 4.4 * 2 + 29.9 = 38.7

(c)    32 < x < 35

z( 32 < z <35 ) = (( 32 - 29.9) / 4.4 < z <( 35 - 29.9) / 4.4) = (0.47 < z <1.16)

3 < z < 4

x(z) = std.deviation * z + mean

x( 3) = 4.4 * 3 + 29.9 = 43.1

x( 4) = 4.4 * 4 + 29.9 = 47.5

(43.1 < x < 47.5)

2)

(d)    2.17 < z

x(z) = std.deviation * z + mean

x( -2.17) = 4.4 * -2.17 + 29.9 = 20.3

5 < x

   z( 5 ) = ( 5 - 29.9) / 4.4 = -5.6

(e)    z < 1.28

x(z) = std.deviation * z + mean

x( 1.28) = 4.4 * 1.28 + 29.9 = 35.5

x < 6

   z( 6 ) = ( 6 - 29.9) / 4.4 = -5.4

(f)    1.99 < z < 1.44

x(z) = std.deviation * z + mean

x( -1.99) = 4.4 * -1.99 + 29.9 = 21.1

x( 1.44) = 4.4 * 1.44 + 29.9 = 36.2

(43.1 < x < 47.5)


7 < x < 8

z( 7 < z < 8 ) = (( 7 - 29.9) / 4.4 < z <( 8 - 29.9) / 4.4) = (-5.2 < z < -4.9)