Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 28.1 kilograms and standard deviation σ = 3.2 kilograms. Let x be the weight of a fawn in kilograms.

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 4.41 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.Yes. This weight is 2.20 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.    No. This weight is 4.41 standard deviations below the mean; 14 kg is a normal weight for a fawn.No. This weight is 4.41 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.No. This weight is 2.20 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.

Answer 1