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Quiz: Introduction to Statistics

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Added on  2021-04-21

Quiz: Introduction to Statistics

   Added on 2021-04-21

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Running head: QUIZ ON STATISTICSQuiz on StatisticsName of the StudentName of the UniversityAuthor Note
Quiz: Introduction to Statistics_1
1QUIZ ON STATISTICSTable of ContentsAnswer to Week 5 Homework...................................................................................................2Explanation of Answer to Week 5 Homework..........................................................................3Answer to Week 6 Homework...................................................................................................6Explanation of Answer to Week 6 Homework..........................................................................8
Quiz: Introduction to Statistics_2
2QUIZ ON STATISTICSAnswer to Week 5 Homework5.2.5 P (64< X< 78) = 0.3439 (Round to four decimal places as needed).5.2.8 The standard deviation is not provided. Hence, the answer cannot be calculated.5.2.11(a) The probability that a randomly selected fertilized egg hatches in less than 20 days is0.1587 (Round to four decimal places as needed).(b) The probability that a randomly selected fertilized egg hatches between 19 and 21 days is= 0.4772(Round to four decimal places as needed).(c)The probability that a randomly selected fertilized egg takes over 22 days to hatch is0.1587(Round to four decimal places as needed).5.2.17(a) Approximately 65.17% of the SAT verbal scores are less than 550 (Round to two decimalplaces as needed).(b) I would expect 433 SAT verbal scores to be greater than 525, when 1000 SAT verbalscores are randomly selected.5.3.33The picture is not given properly. Hence, this question cannot be answered.5.3.34(a) The waiting time that represents the 90th percentile is 1935 days [Round to nearest integeras needed].(b) The waiting time that represents the first quartile is 1526 days [Round to nearest integeras needed].
Quiz: Introduction to Statistics_3
3QUIZ ON STATISTICS5.4.1μ ́x = 90 and σ ́x= 35.4.27-T The probability that the mean salary of the sample is less than $60500 is 0.0031(Round to four decimal places as needed).5.4.31 The probability that the mean height for the sample is greater than 64 inches is 0.2389(Round to four decimal places as needed).5.4.38 Assuming the manufacturer’s claim is correct, the probability that the mean of thesample is 50897 miles or less 0.1271(Round to four decimal places as needed).Explanation of Answer to Week 5 Homework5.2.5 The random variable X follows a normal distribution with mean, μ= 80 and σ=5. Then,the random variable Z=X805 follows a standard normal distribution.The required probability is P (64<X<78) =P (64805<Z<78805); where Z follows normaldistribution with mean=0, variance=1.Hence, the probability is P (-3.2<Z<-0.4) =P(X0.4) - P(x3.2) =0.34395.2.8 The standard deviation is not provided. Hence, the answer cannot be calculated.5.2.11 Incubation time is normally distributed with mean 21 days and standard deviation1day.(a) Probability that a randomly selected fertilized egg hatches in less than 20 days can bewritten as P(X<20).P(X<20) =P (x211<20211 ); where X follows normal distribution with mean=21 days andvariance =1 day
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