Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Fawns between 1 and 5 months old have a body weight that is approximately normal

ID: 3062878 • Letter: F

Question

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean = 28.4 kilograms and standard deviation = 3.2 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x intervals to z intervals. (Round your answers to two decimal places.) (a) x < 30 z < (b) 19 < x < z (c) 32 < x < 35 < z < Convert the following z intervals to x intervals. (Round your answers to one decimal place.) (d) 2.17 < z < x (e) z < 1.28 x < (f) 1.99 < z < 1.44 < x < (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 4.50 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. Yes. This weight is 2.25 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. No. This weight is 4.50 standard deviations below the mean; 14 kg is a normal weight for a fawn. No. This weight is 4.50 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. No. This weight is 2.25 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 z of 0. It would have a large positive z, such as 3. 15. –/1 points Ask Your Teacher My Notes Question Part Points Submissions Used 1 –/1 0/2 Total –/1 Suppose 15% of the area under the standard normal curve lies to the right of a specific Z score. Is that Z score positive or negative? positive negative

Explanation / Answer

a)

a)

b)

c)

d)

as x =mean+z*Std deviaiton

21.46<x

e)

x<32.50

f) 22.03<x<33.01

g)Yes. This weight is 4.50 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.

h) It would have a large positive z, such as 3

i)  positive

for normal distribution z score =(X-)/ here mean=       = 28.4 std deviation   == 3.20000