The following figure is a normal curve that represents the approximate weights, in pounds, of adult female cats of a certain breed. Area = 0.3577 Area = 0.0557 8.5 11.0 Part: 0 / 3 Part 1 of 3 (a) Find the proportion of cats who weigh more than 8.5 pounds. The proportion of cats who weigh more than 8.5 pounds is
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![The following figure is a normal curve that represents the approximate weights, in pounds, of adult female cats of a certain breed.
Area = 0.3577
Area = 0.0557
8.5
11.0
Part: 0 / 3
Part 1 of 3
(a) Find the proportion of cats who weigh more than 8.5 pounds.
The proportion of cats who weigh more than 8.5 pounds is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F231fae22-ac62-4382-9eb0-95d912af77d1%2Ff53f9044-d19a-41fa-bbd9-83bfc998af62%2Fbkc3l17_processed.png&w=3840&q=75)
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- Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05 H0: μd=0 H1: μd≠0 T statistic: t = 1.053, P value = 0.327 Degrees of Freedom n-1 = 8-1= 7 critical value 2.3646 If t is less than 2.3646, or greater than 2.3646, reject the null hypothesis Level of significance: α=0.05 question: what a type 1 error and type 2 error would mean. Is it possible that we could have committed a type 2 error in conducting the testGlaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q) If a participant has the same corneal thickness in their normal eye as the eye with Glaucoma, what would be the value for difference: measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma.Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Q)Conduct a hypothesis test to determine if there is sufficient evidence to conclude that corneal thickness is different in normal eyes compared to eyes with glaucoma? Write up your results using the 8 steps.
- Glaucoma is a leading cause of blindness in the United States, N. Ehlers measured the difference in corneal thickness (in microns) between the two eyes of eight patients. Each patient had one eye that had glaucoma and one eye that was normal. The difference was measured as the corneal thickness of normal eye – corneal thickness of eye with Glaucoma. Corneal thickness is important because it can mask an accurate reading of eye pressure. Use ? = .05. Hypothesis: H0: μd=0Ha: μd≠0 Using output Test statistics : t=0.134P value=0.897 Degrees of freedom (df): df=7 Level of significance: α=0.05 Decision: P value > 0.05 thus we fails to reject null hypothesis. Question a)Write a report summarizing your findings. When writing the report consider that medical staff estimate that a difference of 4.5 microns or more could impact on their ability to interpret eye pressure correctly. b) Define for the hypothesis stated in part b) what a type 1 error and type 2 error would mean. Is it possible…Given the z-score value z = -0.12 and z = 2.63 (under normal curve) A. Find the area of z = -0.12 B. Find the area of z = 2.63 C. Find the area between z = -0.12 and z = 2.63The blood pressure in millimeters was measured for a large sample of people. The average pressure is 140 mm, and the SD of the measurements is 20 mm. The histogram looks reasonably like a normal curv Use the normal curve to estimate the following percentages. Choose the answer that is closest to being correct. A 19.1% B 30.9% C 38.2% D 69.1% E 68.27% A The percentage of people with blood pressure between 130 and 150 mm. B The percentage of people with blood pressure between 140 and 150 mm. D The percentage of people with blood pressure over 150 mm.
- The length of a 13-year-old female's upper arm is approximately normally distributed with mean μ = 34.6 cm and a standard deviation of o=2.4 cm. Complete parts (a) through (c) below. (a) Draw a normal curve with the parameters labeled. Choose the correct graph below. A. O C. A. A -4.8 -2.4 2.4 4.8 7.2 C. -7.2 -X -103.8 -69.2 -34.6 0 34.6 69.2 103.8 I -X B. Q D. (b) Shade the region that represents the proportion of 13-year-old females whose upper arm length is less than 32 cm. The horizontal axis tick marks are the same as those in part (a). Choose the correct graph below. B. I I D. 27.4 29.8 32.2 34.6 -X 37 39.4 41.8 -X -101.4 -66.8 -32.2 2.4 37 71.6 106.2 -XSuppose that a customer is purchasing a car. He conducts an experiment in which he puts 10 gallons of gas in the car and drives it until it runs out of gas. He conducts this experiment 15 times on each car and records the number of miles driven. Median for Car 2 M = 250 mi / 10 gal (Type an integer or decimal rounded to one decimal place as needed.) Full data setO Range for Car 1 R=O mi / 10 gal (Type an integer or decimal rounded to one decimal place as needed.) Car 1 245 242 215 244 220 265 291 160 284 251 169 319 257 306 267 Range for Car 2 R=O mi / 10 gal (Type an integer or decimal rounded to one decimal place as needed.) Car 2 230 206 214 236 242 256 245 255 241 264 276 252 250 251 257Suppose the lengths of human pregnancies are normally distributed with =266 days and o-16 days. Complete parts (a) and (b) below (a) The figure to the right represents the normal curve with u=266 days and e-16 days. The area to the right of X-300 is 0.0168. Provide two interpretations of this area Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice (Type integers or decimals) OA. The proportion of human pregnancies that last more than OB. The proportion of human pregnancies that last less than days is days is Select the correct choice below and fill in the answer boxes to complete your choice days is Provide a second interpretation of the area using the given values (Type integers or decimals) OA. The probability that a randomly selected human pregnancy lasts less than OB. The probability that a randomly selected human pregnancy lasts more than A 4 days is
- Suppose the lengths of human pregnancies are normally distributed with u= 266 days and o = 16 days. Complete parts (a) and (b) below. (a) The figure to the right represents the normal curve with u= 266 days and o = 16 days. The area to the right of X = 300 is 0.0168. Provide two interpretations of this area. Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals.) O A. The proportion of human pregnancies that last less than days is 266 300 O B. The proportion of human pregnancies that last more than days is Provide a second interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals.) O A. The probability that a randomly selected human pregnancy lasts more than days is O B. The probability that a randomly selected human pregnancy lasts less than days isThe following figure is a normal curve that represents the approximate heights, in inches, of adult women in the United States. Area = 0.1029 %3D Area = 0.2036 61 67 Part: 0/ 3 Part 1 of 3 (a) Find the proportion of women who are more than 61 inches tall. The proportion of women who are more than 61 inches tall is(b) The figure to the right represents the normal curve with u= 266 days and o = 16 days. The area between x = 285 and x= 300 is 0.1007. Provide two interpretations of this area. Provide one interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals. Use ascending order.) O A. The proportion of human pregnancies that last less than or more than days is O B. The proportion of human pregnancies that last between and days is X 266 285 300 Provide a second interpretation of the area using the given values. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals. Use ascending order.) O A. The probability that a randomly selected human pregnancy lasts between and days is O B. The probability that a randomly selected human pregnancy lasts less than or more than days is
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