Trusted by +2 million users,

1000+ happy students everyday

1000+ happy students everyday

Showing pages 1 to 4 of 13 pages

Running head: QUIZ ON STATISTICSQuiz on StatisticsName of the StudentName of the UniversityAuthor Note

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

2QUIZ ON STATISTICSAnswer to Week 5 Homework5.2.5P (64< X< 78) =0.3439(Round to four decimal places as needed).5.2.8The 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)Approximately65.17%of the SAT verbal scores are less than 550 (Round to two decimalplaces as needed).(b)I would expect433SAT 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 90thpercentile is1935 days[Round to nearest integeras needed].(b)The waiting time that represents the first quartile is1526 days[Round to nearest integeras needed].

3QUIZ ON STATISTICS5.4.1μ́x=90andσ́x=35.4.27-TThe probability that the mean salary of the sample is less than $60500 is0.0031(Round to four decimal places as needed).5.4.31The probability that the mean height for the sample is greater than 64 inches is0.2389(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 less0.1271(Round to four decimal places as needed).Explanation of Answer to Week 5 Homework5.2.5The random variable X follows a normal distribution with mean,μ= 80 andσ=5. Then,the random variable Z=X−805follows a standard normal distribution.The required probability is P (64<X<78) =P (64−805<Z<78−805); where Z follows normaldistribution with mean=0, variance=1.Hence, the probability is P (-3.2<Z<-0.4) =P(X≤−0.4) - P(x≤−3.2)=0.34395.2.8The standard deviation is not provided. Hence, the answer cannot be calculated.5.2.11Incubation 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 (x−211<20−211); where X follows normal distribution with mean=21 days andvariance =1 day

**You’re reading a preview**

To View Complete Document

Click the button to download

Subscribe to our plans