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.) -2.17 < z (Fill in the blank. A blank is represented by (d) (e) z< 1.28 -1.99 < z< 1.44 (Fill in the blanks. A blank is represented by There are two answer blanks.) (f) 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 3.21 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. Yes. This weight is 1.61 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. No. This weight is 3.21 standard deviations below the mean; 14 kg is a normal weight for a favwn. O No. This weight is 3.21 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. No. This weight is 1.61 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 large positive z, such as 3. It would have a negative z, such as -2. It would have a z of 0. Fawns between 1 and 5 months old have a body weight that is approximately normally distributed vith mean u = 26.2 kilograms and standard deviation o = 3.8 kilograms. Let x be the weight of a fawn in kilograms. The Standard Normal Distribution (u = 0, o - 1) -2 -5 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 z<84 19 < x (Fill in the blank. A blank is represented by (Ь) |-1.89 32 < x < 35 (Fill in the blanks. A blank is represented by (c) 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.) -2.17 < z (Fill in the blank. A blank is represented by (d) (e) z< 1.28 -1.99 < z< 1.44 (Fill in the blanks. A blank is represented by There are two answer blanks.) (f) 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 3.21 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. Yes. This weight is 1.61 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. No. This weight is 3.21 standard deviations below the mean; 14 kg is a normal weight for a favwn. O No. This weight is 3.21 standard deviations above the mean; 14 kg is an unusually high weight for a fawn. No. This weight is 1.61 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 large positive z, such as 3. It would have a negative z, such as -2. It would have a z of 0. Fawns between 1 and 5 months old have a body weight that is approximately normally distributed vith mean u = 26.2 kilograms and standard deviation o = 3.8 kilograms. Let x be the weight of a fawn in kilograms. The Standard Normal Distribution (u = 0, o - 1) -2 -5 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 z<84 19 < x (Fill in the blank. A blank is represented by (Ь) |-1.89 32 < x < 35 (Fill in the blanks. A blank is represented by (c) There are two answer blanks.) first blank second blank
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
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ISBN:9781680331141
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Chapter7: Polynomial Equations And Factoring
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Transcribed Image Text: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.)
-2.17 < z (Fill in the blank. A blank is represented by
(d)
(e) z< 1.28
-1.99 < z< 1.44 (Fill in the blanks. A blank is represented by
There are two answer blanks.)
(f)
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 3.21 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
Yes. This weight is 1.61 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.
No. This weight is 3.21 standard deviations below the mean; 14 kg is a normal weight for a favwn.
O No. This weight is 3.21 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
No. This weight is 1.61 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 large positive z, such as 3.
It would have a negative z, such as -2.
It would have a z of 0.
Transcribed Image Text:Fawns between 1 and 5 months old have a body weight that is approximately normally distributed vith mean u = 26.2 kilograms and standard deviation o = 3.8 kilograms. Let x be the weight of a fawn in kilograms.
The Standard Normal Distribution
(u = 0, o - 1)
-2
-5
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
z<84
19 < x (Fill in the blank. A blank is represented by
(Ь)
|-1.89
32 < x < 35 (Fill in the blanks. A blank is represented by
(c)
There are two answer blanks.)
first blank
second blank
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