MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean u = 29.6 kilograms and standard deviation o = 5.0 kilograms. Let x be the weight of a fawn in kilograms.
The Standard Normal Distribution
u = 0, o = 1)
-2
3
68% of area
95% of area
99.7% of area
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
1x
(b) 19 < x (Fill in the blank. A blank is represented by )
(c) 32 < x < 35 (Fill in the blanks. A blank is represented by There are two answer blanks.)
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 )
(e)
z< 1.28
(f)
-1.99 < z< 1.44 (Fill in the blanks. A blank is represented by
There are two answer blanks.)
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.
O Yes. This weight is 3.12 standard deviations below the mean; 14 kg is an unusually lovw weight for a fawn.
O Yes. This wreight is 1.56 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
O No. This weight is 3.12 standard deviations below the mean; 14 kg is a normal weight for a fawn.
O No. This weight is 3.12 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
O No. This weight is 1.56 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
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Transcribed Image Text:Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean u = 29.6 kilograms and standard deviation o = 5.0 kilograms. Let x be the weight of a fawn in kilograms. The Standard Normal Distribution u = 0, o = 1) -2 3 68% of area 95% of area 99.7% of area 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 1x (b) 19 < x (Fill in the blank. A blank is represented by ) (c) 32 < x < 35 (Fill in the blanks. A blank is represented by There are two answer blanks.) 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 ) (e) z< 1.28 (f) -1.99 < z< 1.44 (Fill in the blanks. A blank is represented by There are two answer blanks.) 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. O Yes. This weight is 3.12 standard deviations below the mean; 14 kg is an unusually lovw weight for a fawn. O Yes. This wreight is 1.56 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. O No. This weight is 3.12 standard deviations below the mean; 14 kg is a normal weight for a fawn. O No. This weight is 3.12 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. O No. This weight is 1.56 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
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