Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 28.0 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 4.38 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. Yes. This weight is 2.19 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. No. This weight is 4.38 standard deviations below the mean; 14 kg is a normal weight for a fawn. No. This weight is 4.38 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. No. This weight is 2.19 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 negative z, such as −2. It would have a large positive z, such as 3. It would have a z of 0.