Question

Question 8 (1 point) A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 28.7 and 30 hours studying? Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage. Your Answer: Answer Question 9 (1 point) Electricity bills in a certain city have mean $ 107.78. Assume the bills are normally! distributed with standard deviation $ 14.82. A sample of 74 bills was selected for an audit. Find the 54 percentile for the sample mean. Write only a number as your answer. Round to two decimal places (for example: 42.81). Do not write any units. Your Answer:

Answer 1

solutions Question : 6 A Gorley among fadhmen ata Corlain Conilcorcityproevaled : Rezcuired informations i@gion by :15 Mean= = 05 Glancoid deviation = Sample (size = 36 e Speed Calcabre i What is the probability that the average time spend it kle have to obtained the probability that the average # is) Eleres »probability is PCOB. + < X < 30] 5p 38.7-84 - 1 an <tes <30-01 cirocras op [ <2443) plot 2 2 2 e os PC1.418 <<<3) -> p<Zc0- przch ciB]

-> pit col- P(Z<l.608] -> 6.9+ ] -[0.98061 -) (6. oub6] Therefore, Ragurired probabiling is (0.0166) Quastin - Electricity bills in a Cortain CHH balle mean 107-48 Rezuired data ils ll= 107.48 = Mean c=14.82 =tas dord deviation. Komple Gize=74 Calculate - Find the shit percentile Candle meant- the 54 percentile of Comple mean is, 66 have to obtained Ekre, sample nean is = 541:42 PCT < PC

PPCZC -0.541 -> 2=770341) 2 - 0 1004 1-4 - 0.100 =0.1004 = (x - 107.48] nu.82) Cra - -0 1004 -> (8-107.48] 142014 -> -107.78] = [o.looux 1.70979) Tye 10+.+8) = 0.14096 8 F = 107. +8+0.172968 107.95) Therefore, Sample Mean is ya 101.95