Qnt_561_Week_1

Assignment Week 1 QNT/561 Chapter 3 80. Creek Ratz is a very popular restaurant located along the coast of northern Florida. They serve a variety of steak and seafood dinners. During the summer beach season, they do not take reservations or accept “call ahead” seating. Management of the restaurant is concerned with the time a patron must wait before being seated for dinner. Listed below is the wait time, in minutes, for the 25 tables seated last Saturday night. |23 | |24 | |24 | |25 | |27 | |27 | |28 | |28 | |28 | |29 | |34 | |37 | |39 | |40 | |44 | |44 | |45 | |50 | |51 | |56 | |61 | |61 | |64 | |65 | |67 | |1021 | |40.84 | a. Explain why the times are a population. El tiempo de espera es y se tratan como una populación, ya que al tener una gran cantidad de clientes esperando puede crear no tan solo molestia para los clientes sino también puede crear una perdida para la gerencia de los restaurantes. b. Find the mean and median of the times. Mean= 40.84 Median= 39 c. Find the range and the standard deviation of the times Range: 44 Standard Deviation: 14.8 82. The following frequency distribution reports the electricity cost for a sample of 50 two bedroom apartments in Albuquerque, New Mexico during the month of May last year. | |Midpoint | |Frecuency | |M-X |(M-X)2 |F(M-X)2 | |80 |100 |90 | |3 | |270 |-51.2 |2621.4 |9525.8 | |100 |120 |110 | |8 | |880 |-31.2 |973.4 |7787.5 | |120 |140 |130 | |12 | |1560 |-11.2 |125.4 |1505.3 | |140 |160 |150 | |16 | |2400 |8.8 |77.4 |1239.0 | |160 |180 |170 | |7 | |1190 |28.8 |829.4 |5806.1 | |180 |200 |190 | |4 | |760 |48.8 |2381.4 |9525.8 | | | | |total |50 | |7060 | | |35389.4 | | | | | |Mean= |141.2 | | | |722.2 | | a. Estimate the mean cost. Mean=141.2 b. Estimate the standard deviation. Standard Deviation=26.9 c. Use the Empirical Rule to estimate the proportion of costs within two standard deviations of the mean. What are these limits' 87.5 limite bajo 194.9 limite alto 87. Refer to the Real Estate data, which reports information on homes sold in the Denver, Colorado, area last year. Price |Bedrooms |Size |Pool |Distance |Twnship |Garage |Baths | |125 |8 |2900 |0 |9 |4 |1 |2 | |125.9 |8 |2600 |0 |9 |4 |1 |2 | |139.9 |7 |2400 |0 |8 |4 |1 |2 | |147.4 |7 |2400 |0 |13 |3 |1 |3 | |154.3 |6 |1700 |1 |12 |1 |0 |2 | |155.4 |6 |2100 |1 |14 |2 |1 |2.5 | |164.1 |6 |2100 |1 |15 |4 |1 |2 | |166.2 |6 |2200 |1 |14 |1 |0 |2 | |166.5 |6 |1900 |1 |15 |4 |1 |2 | |171.6 |6 |2700 |0 |7 |4 |1 |2 | |172.4 |6 |2300 |0 |7 |3 |1 |3 | |172.7 |6 |2500 |0 |7 |4 |1 |2 | |173.1 |6 |2000 |0 |21 |3 |1 |3 | |173.6 |6 |2700 |0 |15 |3 |1 |2 | |173.6 |6 |2400 |0 |11 |5 |1 |3 | |175 |6 |2600 |0 |7 |5 |1 |2.5 | |175.6 |6 |2100 |0 |11 |5 |1 |3 | |176 |6 |2500 |0 |15 |3 |1 |2 | |176.3 |5 |2500 |1 |21 |4 |0 |2.5 | |177.1 |5 |2300 |0 |11 |5 |1 |3 | |179 |5 |2300 |1 |21 |4 |0 |2.5 | |179 |5 |2300 |0 |20 |5 |0 |1.5 | |180.4 |5 |2100 |0 |20 |5 |0 |1.5 | |182.4 |5 |2200 |1 |13 |2 |1 |2 | |182.7 |5 |2200 |0 |20 |3 |1 |2 | |186.7 |5 |2400 |0 |16 |2 |1 |2 | |187 |5 |2200 |0 |8 |5 |1 |3 | |188.1 |5 |2200 |1 |11 |4 |1 |2.5 | |188.3 |5 |2200 |0 |16 |2 |1 |3 | |188.3 |4 |2200 |1 |16 |3 |0 |2 | |188.3 |4 |2400 |1 |16 |3 |0 |2 | |188.3 |4 |2300 |0 |19 |4 |0 |2 | |189.4 |4 |2100 |0 |9 |4 |1 |2 | |190.9 |4 |2100 |1 |14 |2 |1 |2.5 | |192.2 |4 |2300 |1 |24 |4 |1 |2 | |192.6 |4 |2200 |0 |15 |1 |1 |2 | |192.9 |4 |2100 |0 |19 |4 |0 |2 | |194.4 |4 |2000 |1 |14 |4 |0 |2.5 | |198.3 |4 |2200 |0 |24 |1 |1 |2 | |198.9 |4 |1900 |1 |14 |2 |1 |2.5 | |199 |4 |2100 |1 |19 |1 |1 |1.5 | |199.8 |4 |2200 |1 |19 |2 |1 |2 | |205.1 |4 |2300 |0 |19 |2 |0 |2 | |206 |4 |2300 |0 |18 |3 |1 |2 | |207.1 |4 |2900 |0 |20 |5 |0 |1.5 | |207.5 |4 |2900 |0 |8 |4 |1 |2 | |207.5 |4 |2600 |1 |17 |5 |1 |2 | |209 |4 |2600 |0 |13 |4 |1 |2 | |209.3 |4 |2100 |0 |18 |3 |1 |2 | |209.3 |4 |2600 |0 |8 |4 |1 |2 | |209.7 |4 |2500 |1 |15 |3 |1 |2 | |209.7 |4 |2300 |1 |17 |5 |1 |2 | |213.6 |4 |2300 |0 |14 |3 |1 |2 | |216 |4 |2400 |0 |13 |4 |1 |2 | |216.8 |4 |2100 |0 |14 |3 |1 |2 | |217.8 |3 |2000 |1 |16 |2 |1 |2 | |220.9 |3 |1600 |1 |19 |3 |0 |2.5 | |221.1 |3 |2000 |1 |16 |4 |0 |2 | |221.5 |3 |2200 |0 |23 |3 |0 |2 | |222.1 |3 |2400 |0 |8 |4 |1 |2 | |224 |3 |2400 |1 |10 |4 |1 |2 | |224.8 |3 |2200 |1 |18 |3 |1 |2 | |227.1 |3 |2500 |1 |18 |1 |0 |1.5 | |227.1 |3 |2200 |1 |10 |4 |1 |2 | |228.4 |3 |2100 |0 |19 |3 |1 |2 | |232.2 |3 |2000 |1 |20 |4 |0 |2 | |233 |3 |2100 |1 |9 |3 |0 |1.5 | |234 |3 |2100 |1 |10 |2 |0 |2 | |236.4 |3 |2200 |0 |15 |1 |1 |2 | |236.8 |3 |2500 |0 |12 |3 |0 |2 | |240 |3 |2300 |1 |18 |1 |0 |1.5 | |242.1 |3 |1900 |1 |6 |1 |1 |2 | |243.7 |3 |2200 |0 |17 |1 |1 |2.5 | |244.6 |3 |2300 |0 |17 |5 |1 |1.5 | |245.4 |3 |1900 |1 |16 |1 |1 |1.5 | |246 |3 |2200 |0 |14 |3 |1 |1.5 | |246.1 |3 |2300 |0 |12 |3 |0 |2 | |247.7 |3 |1900 |0 |12 |2 |1 |2 | |251.4 |3 |2300 |0 |16 |2 |1 |2 | |252.3 |3 |2100 |0 |16 |2 |1 |2 | |253.2 |3 |2400 |1 |21 |2 |1 |3 | |254.3 |2 |1900 |0 |18 |4 |0 |1.5 | |257.2 |2 |2400 |0 |28 |1 |0 |1.5 | |263.1 |2 |2100 |0 |28 |1 |0 |1.5 | |263.2 |2 |2000 |0 |13 |2 |0 |2 | |266.6 |2 |2200 |1 |21 |5 |1 |1.5 | |269.2 |2 |2500 |0 |11 |3 |0 |2 | |269.9 |2 |2000 |1 |17 |3 |0 |2 | |270.8 |2 |1900 |0 |10 |5 |1 |2 | |271.8 |2 |2000 |0 |11 |5 |0 |2 | |273.2 |2 |1900 |0 |26 |4 |0 |2 | |281.3 |2 |1900 |0 |8 |4 |1 |1.5 | |289.8 |2 |2400 |0 |16 |2 |0 |2.5 | |292.4 |2 |2300 |0 |11 |3 |0 |2 | |293.7 |2 |2000 |0 |11 |5 |1 |2 | |294 |2 |1700 |0 |8 |4 |1 |1.5 | |294.3 |2 |2200 |0 |16 |2 |0 |2.5 | |294.5 |2 |2300 |1 |12 |1 |1 |2 | |307.8 |2 |2100 |0 |9 |5 |1 |2 | |310.8 |2 |1700 |0 |19 |3 |1 |2 | |312.1 |2 |2100 |1 |12 |1 |1 |2 | |312.1 |2 |2300 |0 |9 |2 |1 |2.5 | |326.3 |2 |2100 |0 |9 |4 |1 |2 | |327.2 |2 |2100 |0 |9 |2 |1 |2.5 | |345.3 |2 |2100 |0 |13 |2 |1 |2.5 | |23215.8 | |233500 | | | | | | | a. Select the variable selling price. 1. Find the mean, median, and the standard deviation. Mean=221.1 Median= 213.6 Standard Deviation= 47.1 2. Write a brief summary of the distribution of selling prices. La media para el costo de venta es alrededor de $221,100, y su mediana es más baja: $213,600. La desviación estándar es de $47,100. b. Select the variable referring to the area of the home in square feet. 1. Find the mean, median, and the standard deviation. Mean= 2223.8 Median=2,200 Standard Deviation=248.7 2. Write a brief summary of the distribution of the area of homes. La media para las medidas cuadradas es de 2224 sq ft, la mediana es de: 2200 sq ft. La desviación estándar es de 249 sq ft. Chapter 5 34. P (A1) =.20, P (A2) =.40, and P (A3) =.40. P (B1I A1) =.05, and P (B1IA3) =.10. USE BAYES THEOREM TO DETERMINE P (A3IB1). P (A3I B1)= P(A3) P(B1IA3)= (.40)(.10)=.04 36. Dr. Stallter has been teaching basic statistics for many years. She knows that 80 percentof the students will complete the assigned problems. She has also determined that among those who do their assignments, 90 percent will pass the course. Among those students who do not do their homework, 60 percent will pass. Mike Fishbaugh took statistics last semester from Dr. Stallter and received a passing grade. What is the probability that he completed the assignments' P= .80(.90)/ {.20(.60)+ .80 (.90)} P=.72/ .84 P=.86 38. One-fourth of the residents of the Burning Ridge Estates leave their garage doors open when they are away from home. The local chief of police estimates that 5 percent of the garages with open doors will have something stolen, but only 1 percent of those closed will have something stolen. If a garage is robbed, what is the probability the doors were left open' TOTAL DE RESIDENTES QUE DEJARON LAS PUERTAS DE GARAJES ABIERTAS 1/4=25% TOTAL DE RESIDENTES QUE NO DEJARON LAS PUERTAS DE GARAJE ABIERTA 3/4=75% 25% X 5%= .0125 75%X1%=0.0075 .0125-0.0075=.005 La probabilidad de hogares que dejaron las puertas de garajes abiertas es de un .5%. Chapter 6 45. A Tamiami shearing machine is producing 10 percent defective pieces, which is abnormally high. The quality control engineer has been checking the output by almost continuous sampling since the abnormal condition began. What is the probability that in a sample of 10 pieces: a. Exactly 5 will be defective' P(x)= nCx (π)x (1- π)n-x P(5)= 10C5 (0.1)5 (1- .10)10-5 = 252(0.1)5(0.9)5 =0.001488 b. 5 or more will be defective' P(x)= nCx (π)x (1- π)n-x =10C5 (0.1)5 (0.9)5 + 10C6 (0.1)6(0.9)4+10C7(0.1)7(0.9)3+ 10C8(0.1)8(0.9)2+10C9 (0.1)9(0.9)1+10C10 (0.1)10(0.9)0 =252(0.1)5(0.9)5+210(0.1)6(0.9)4+120(0.1)7(0.9)3+45(0.1)8(0.9)2+10(0.1)9(0.9)1+1(0.1)10(0.9)0 =0.00163 62. Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random sample of 200 antennas, find the probability that: a. None of the antennas is defective. = (1 - 0.015)200 = 0.04866829 Chapter 7 42. The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect' [pic] z(29) = (29-32) / 2 = -1.5 z(34) = (34-32) / 2 = 1 What percent of the garages take between 32 hours and 34 hours to erect' prob(0) < z < 1 b. What percent of the garages take between 29 hours and 34 hours to erect' prob(-1.5 < z < 1) -- 77.45% c. What percent of the garages take 28.7 hours or less to erect' z value =(28.7-32)/2 = -1.65) prob(z < -1.65) 4.95% d. Of the garages, 5 percent take how many hours or more to erect' z value = 1.6449 m + σ= 32 + 1.6449(2) =35.2898 hours 45. Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a standard deviation of $420 per year. a. What fraction of the employees cost more than $1,500 per year for dental expenses' z = (1500-1280)/420 z = 0.5238 prob(z > 0.5238) = 0.3002 (3/10) b. What fraction of the employees cost between $1,500 and $2,000 per year' z(2000) = (2000-1280)/420 z = 1.71429 prob(0.5238 < z < 1.71429) = 0.2570 ( 1/4) c. Estimate the percent that did not have any dental expense. z(0) = (0-1280)/420 z = -3.0476 prob(z < -3.0476) = 0.0012 (0.12%) d. What was the cost for the 10 percent of employees who incurred the highest dental expense' z = 1.2816. X=' z = 1.2816 = (x-1280)/420 x-1280 = 420(1.2816) x = 420(1.2816) + 1280 x = 1818.272 > $1818.27