Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean = 27.3 kilograms and standard deviation = 3.6 kilograms. Let x be the weight of a fawn in kilograms.

Standard Normal Distribution

= 0; = 1

-1 to 1 = 68% of area; -2 to 2 = 95% of area; -3 to 3 = 99.7% of area

For parts (a), (b), and (c), convert the x intervals to z intervals. (For each answer, enter a number. Round your answers to two decimal places.)

(a)

x < 30

z < ____

(b) 19 < x (Fill in the blank. A blank is represented by _____.)

_____ < z

(c) 32 < x < 35 (Fill in the blanks. A blank is represented by _____. There are two answer blanks.)

_____ < z < _____

first blank =

second blank =

For parts (d), (e), and (f), convert the z intervals to x intervals. (For each answer, enter a number. Round your answers to one decimal place.) (d)

2.17 < z (Fill in the blank. A blank is represented by _____.)

_____ < x

(e) z< 1.28

x < ______

(f)1.99 <z< 1.44 (Fill in the blanks. A blank is represented by _____. There are two answer blanks.)

_____ <x< _____

first blank =

second blank =

(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 3.69 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
• Yes. This weight is 1.85 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
• No. This weight is 3.69 standard deviations below the mean; 14 kg is a normal weight for a fawn.
• No. This weight is 3.69 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
• No. This weight is 1.85 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 negative z, such as 2.
• It would have a large positive z, such as 3.