- Have any questions?
- +1 (347) 670-5218
- support@nursinghomeworks.org

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.