Answer Happy

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ = 25.8 kilograms and standard deviation = 3.5 kilograms. Let x be the weight of a fawn in kilograms. The Standard Normal Distribution (μ = 0, σ = 1) -3 <Z -2 -1 <>< 0 68% of area 95% of area 99.7% of area (b) 19 < x (Fill in the blank. A blank is represented by first blank second blank 2 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< 3 Z (c) 32 < x < 35 (Fill in the blanks. A blank is represented by There are two answer blanks.)
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 ___________.) (e) z < 1.28 (f) -1.99 < z < 1.44 (Fill in the blanks. A blank is represented by < x < _________ first blank second blank There are two answer blanks.) (g) If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and the figure above. O Yes. This weight is 3.37 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. O Yes. This weight is 1.69 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. O No. This weight is 3.37 standard deviations below the mean; 14 kg is a normal weight for a fawn. O No. This weight is 3.37 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. O No. This weight is 1.69 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. O It would have a negative z, such as -2. O It would have a z of 0. O It would have a large positive z, such as 3.