For the standard normal curve, find the z-score that corresponds to the 90^th pe

Question

For the standard normal curve, find the z-score that corresponds to the 90^th percentile. A) 1.52 B) 0.28 C) 1.28 D) 2.81 Compare the scores: a score of 88 on a test with a mean of 79 and a score of 78 on a test with a mean of 70 A) A score of 75 with a mean of 70 and a standard deviation of 4 is better. B) The two scores are statistically the same. C) You cannot determine which score is better from the given information. D) A score of 75 with a mean of 65 and a standard deviation of 8 is better. SAT scores have a mean of 1026 and a standard deviation of 209. ACT scores have a mean of 20.8 and a standard deviation of 4.8. A student takes both tests while a junior and scores 860 on the SAT and 16 on the ACT. Compare the scores. A) A score of 860 on the SAT test was better. B) A score of 16 on the ACT test was better. C) You cannot determine which score is better from the given information. D) The two scores are statistically the same. In a certain normal distribution, find the standard deviation sigma when mu = 50 and 10.56% of the area lies to the right of 55. A) 2 B) 4 C) 3 D) 5 The lengths of pregnancies are normally distributed with a mean of 264 days and a standard deviation of 15 days. If 36 women are randomly selected, find the probability that they have a mean pregnancy between 264 days and 266 days. A) 0.5517 B) 0.2881 C) 0.7881 D) 0.2119 Assume that the heights of men are normally distributed with a mean of 69.5 inches and a standard deviation of 2.1 inches. If 36 men are randomly selected, find the probability that they have a mean height greater than 70.5 inches. A) 0.9005 B) 9.9671 C) 0.8188 D) 0.0021

Explanation / Answer

14) option C is correct

15) as std deviaiton not known ; option C is correct

16)option A due to better z score

17)for 10.56% rightmost area ; z=1.25

therefore std deviation =(55-50)/1.25 =4

option B

18) std error of mean =std deviation/(n)1/2 =15/(36)1/2 =2.5

therefore P(264<X<266) =P(0<Z<0.8)= 0.7881-0.5 =0.2881

option B

19)

std error of mean =std deviation/(n)1/2 =2.1/(36)1/2 =0.35

therefore P(X>70.5)=P(Z>2.8571)=0.0021

option D

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.