Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean =27.2 kilograms and standard deviation =4.2...

μ = 27.2σ = 4.2x

xz

(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      

zx

(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 3.14 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

Yes. This weight is 1.57 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.    

No. This weight is 3.14 standard deviations below the mean; 14 kg is a normal weight for a fawn.

No. This weight is 3.14 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.

No. This weight is 1.57 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.