*Only need parts D-H*
Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with meanμ = 28.7 kilograms and standard deviationσ = 3.0 kilograms. Letx be the weight of a fawn in kilograms. For parts (a), (b), and (c), convert thex intervals toz intervals. (For each answer, enter a number. Round your answers to two decimal places.)
(a)x <>
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="">
_____ z <>
first blank
second blank
For parts (d), (e), and (f), convert thez intervals tox 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 <>
x
(f) −1.99 z < 1.44="" (fill="" in="" the="" blanks.="" a="" blank="" is="" represented="" by="" _____.="" there="" are="" two="" answer="">
_____ x <>
first blank
second blank
(g) If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain usingz values and the figure above.
Yes. This weight is 3.10 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
Yes. This weight is 1.55 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
No. This weight is 3.10 standard deviations below the mean; 14 kg is a normal weight for a fawn.
No. This weight is 3.10 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
No. This weight is 1.55 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 thez value for the weight of the fawn will be close to 0, −2, or 3? Explain.
It would have a negativez, such as −2.
It would have az of 0.
It would have a large positivez, such as 3.