Math 302 Week Four Test 1 03 2024

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American Public University *

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302

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Statistics

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Apr 3, 2024

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Question 1 1/1point Find the area under the standard normal distribution to the left of z = -1.35. Round answer to 4 decimal places. Answer: ___0.0885 _ v w Hide question 1 feedback In Excel, =NORM.S.DIST(-1.35,TRUE) Question 2 1/1point Find P(Z < -1.45). Round answer to 4 decimal places. Answer: __0.0735 __ v v Hide question 2 feedback In Excel, =NORM.S.DIST(-1.45TRUE) Question 3 1/1point

The lngth of 3 humsnpregnancy s narmallydistrbuted with 3 mean of 270 days with 2 standard deviation o 8 days. How many days would pegnancy 15t forth shortst 15%? Round answer to 2 decimal psces. Answr: —an_ v v Hide question 3 feedback InEscal NORMINVO15.2708) Questions 1/ 1point Whch sy o i dos e gogh s Rt isriion Question's 1/ 1point The costof uneade gssoline inth By Area ance followed 3 norml istibution with 3 mean of $4.74 nd 3 standard deviaton of $0.16.Twenty- ight gas staion from the Bay ara are andoy chosen. We ae nerested

i the sveragecost o gasline for the 28 gas statons. Find the exact probabity thatthe aversge price for 28 gas satons i fess than $4.69. 0000 o~ oo oosis o801 v Hide question s feedback NewSD = 14/5QRT(28) 0030237 Plx < .69)In Excel, sNORMDIST(4 69.474,0030237.TRUE) Questions 1/1point The sversge et of 3 certainnew cll phane s 3.4 years. The manfacturerwil relace any cll phone falig witinthce years of the date of purchase. TheIfetim of thes call phones is known 1o fllow 3n xpanentil disrbuton. 8% of these phone st how long i years? assss PREL azae2 20794 v ide question 6 feedback For 8% use 631 the cquation and the ate of decay s /3.4

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Question? 0/1point The sversge et of 3 certainnews cll phane s 4.2 years. The manfacturerwil relace any cell phone falig witinthce years of the date of purchase. The Ifetime of thes call phones is known to fllow 3n ponentil distrbution. The decay rte s oozt 07619 %0z oeser v Hide question 7 feedback waz Questions 1/1point “The e of aneectric companent h an exponentisl distrbuton with s mean. 0183 years What i th probabity that a randomly selected on such component has e more than & years? Answer: (Round o 4 decimal places) Ry = Hide question s feedback P8 Pic<n)

=1-EXPON.DIST(8,1/8.9.TRUE) Question9 1/1point The caller times at a customer service center has an exponential distribution ‘with an average of 22 seconds. Find the probability that a randomly selected call time will be less than 30 seconds? (Round to 4 decimal places) Answer: 07443 v v Hide question 9 feedback Plx < 30) In Excel, =EXPON.DIST(30,1/22TRUE) Question 10 1/1point Alocal pizza restaurant delivery time has a uniform distribution over 10 to 75 minutes. What is the probability that the pizza delivery time is more than 30 minutes on a given day? Answer: (Round to four decimal places) 06923 v w Hide question 10 feedback Interval goes from 10 < x < 75 Plx >30) = (75-30) * 75 — 10

Question 11 1/1point ‘The waiting time in line at an ice cream shop has a uniform distribution between 3 and 14 minutes. What is the 75th percentile of this distribution? (Recall: The 75th percentile divides the distribution into 2 parts so that 75% of area is to the left of 75th percentile) minutes Answer: (Round answer to two decimal places) ___1125 _ v w Hide question 11 feedback Interval goes from 3 < x < 14. 75th percentile use .75 PIX <x)= 75 75+ z—3 14-3 75 11=x-3 1125 Question 12 1/1 point ‘The waiting time for a table at a busy restaurant has a uiform distribution between 15 and 45 minutes. What is the 90th percentile of this distribution? (Recall: The 90th percentile divides the distribution into 2 parts so that 90% of area is to the left of 90th percentile) minutes Answer: (Round answer to one decimal place) __42 v Hide question 12 feedback

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Interval goes from 15 < x < 45. 90th percentile use .90 POX <X 90 z—-15 45 —15 90°30 27=x-15 Question 13 1/1point ‘The waiting time for a bus has a uniform distribution between 2 and 11 minutes. What is the 75th percentile of this distribution? ______ minutes Answer: (Round answer to two decimal places) __875__ v/ w Hide question 13 feedback PX<x)= 75 75= -2 1-2

875=x Question 14 1/1point Alocal pizza restaurant delivery time has a uniform distribution over 10 to 75 minutes. What is the probability that the pizza delivery time is more than 45 minutes on a given day? Answer: (Round to four decimal places) 04615 v w Hide question 14 feedback Interval goes from 10 < x < 75 Pix > 45) = (75 45) =L 75— 10 Question 15 1/1point ‘The waiting time for an Uber has a uniform distribution between 5 and 37 minutes. What is the probability that the waiting time for this Uber is less than 13 minutes on a given day? Answer: (Round to two decimal places) 02 v v Hide question 15 feedback Interval goes from 5 < x < 37 Plx<13)=(13-5)"

375 Question 16 0/1point ‘The average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.38 ounces. If a random sample of eighty 16-ounce beverage cans are selected, what is the probability that mean of this sample is less than 16.1 ounces of beverage? Answer: (round to 4 decimal places) __00301 _ x (0.0298, .0298) v Hide question 16 feedback New SD = .38/SQRT(80) = 0.042485 Plx < 16.1) in Excel =NORM.DIST(16.1,16.18,0.042485 TRUE) Question 17 1/1point ‘The average credit card debt for college seniors is $22,199 with a standard deviation of $5300. What is the probability that a sample of 30 seniors owes 2 mean of more than $20,200? Round answer to 4 decimal places. Answer: 09806 v v Hide question 17 feedback New SD = 5300/SQRT(30) = 967.6432 Plx > 20200}, using Excel NORM.DIST(20200,22199,967.6432.TRUE) Question 18 0/1point

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The average amount of a beverage in randomly selected 16-ounce beverage can s 16,18 ounces with a standard deviation of 0.4 ounces. If a random sample of sixty-five 16-ounce beverage cans are selected, what s the probabilty that the mean of this sample s less than 16.1 ounces of beverage? Answer: round to 4 decimal places) __00537 __ x (0.0534,.0534) v Hide question 18 feedback New SD = .4/SQRT(65) = 0049614 Plx < 16.1) in Excel =NORM.DIST(16.1,16.18,0.049614 TRUE) Question 19 1/1point The average amount of a beverage in randomly selected 16-ounce beverage can s 15,85 ounces with a standard deviation of 0.3 ounces. If a random sample of thirty-five 16-ounce beverage cans are selected, what is the probabilty that the mean of this sample s less than 15.7 ounces of beverage? Answer: (round to 4 decimal places) 00015 _ v Hide question 19 feedback NewSD =, /SQRT(3S) = 0050709 P(x <15.7), in Excel =NORM.DIST(15.7,15.85,0.050709,TRUE) Question 20 1/1point “The final exam grade of a satistics class has a skewed distribution with mean f 81.2 and standard deviation of 6.95. If a random sample of 42 students selected from this class, then what i the probabilit that the average final

‘exam grade of this sample is between 75 and 807 Answer: (round to 4 decimal places) _ 01316 __ v v Hide question 20 feedback NewSD = 95/SQRT(42) = 1.072408 P(75 < x < 80), in Excel =NORM.DIST(80,81.2,1.072408.TRUE)- NORM.DIST(75,81.2,1.072408 TRUE)