wns between 1 and 5 months old have a body weight that is approximately normally distributed with mean=29.9kilograms and standard deviation=3.3kilograms. Letxbe the weight of a fawn in kilograms.For parts (a), (b), and (c), convert thexintervals tozintervals. (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 thezintervals toxintervals. (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 usingzvalues and the figure above.

Yes. This weight is 4.82 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

Yes. This weight is 2.41 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

No. This weight is 4.82 standard deviations below the mean; 14 kg is a normal weight for a fawn.

No. This weight is 4.82 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.

No. This weight is 2.41 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 thezvalue for the weight of the fawn will be close to 0,2, or 3? Explain.

It would have a large positivez, such as 3.

It would have azof 0.

It would have a negativez, such as2.