Time left 0:11:58Quiz navigation4Queston 3A set of 120 patients visiting a clinic were classified as followingNot yetFinish attempt.Male ( M ) Female (answeredFMarked out of20301.00With eye glasses( G )Suppose that one patient is selected randomly from this set, then the probability that thePRag question6010Without eye glasses( G )selected patient is male given that the patient is with eye glassesO a 2/5Ob. none of theseO. 1/6Od. 1/4 هذه المادة - العربية )ar( -مقر راتي -AUG-Moodle-83الوقت المتبقي 1 0:12:1A set of 120 patients visiting a clinic were classified as following:ule ube ueMale ( M ) Female ( F )1.00 e dall2030With eye glasses( G )Without eye glasses( G )Suppose that one patient isLia alc PJljuall6010selected randomly from this set, then the probability that the selected patient is male:a O2/380.cO1/3d Onone of these

Question 1 : Correct answer is (a) 2/5

A set of 120 patients visiting a cinic were classified as following Male (M) Female ( F ) 20 With eye glasses(G ) Without eye glasses( G 30 Suppose that one patient is selected randomly from this se 60 10 selected patient is male given that the patient is with eye glasses Oa 25 Ob. none of these Oc 1/6 Od. 1/4

A set of 120 patients visiting a cinic were classified as following Male (M) Female ( F ) 20 With eye glasses(G ) Without eye glasses( G 30 Suppose that one patient is selected randomly from this se 60 10 selected patient is male given that the patient is with eye glasses Oa 25 Ob. none of these Oc 1/6 Od. 1/4

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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3. An integer is randomly chosen from the set 1, 2,..., 100. If this integer is divisible by 2 we let Y = 0, and if it is not divisible by 2 but is divisible by 3 then Y = 1. We let Y = 2 in all other cases. Find mean and variance of Y.
The third worksheet labeled sample B is a simple random sample with replacement, with seven observations in the sample, from some population. The index i in the first column is just an integer name for each observation. The specific values x in the second column are measures of a random variable X distributed in the population. Now suppose this random variable X is known to have a Normal distribution in the sampled population, BUT you do not know the population mean or the population standard deviation of that Normal distribution: Both must be estimated from the sample as you just did. Compute the lower bound of an 80% confidence interval for the population mean, using the appropriate sample estimates and techniques appropriate to the knowledge situation described above. i xi 1 74.2 2 72.79 3 70.35 4 74.37 5 73.34 6 78.96 7 77.99
A new process has been designed to make ceramic tiles. The goal is to have no more than 5% of the tiles be nonconforming due to surface defects. A random sample of 1000 tiles is inspected. Let X be the number of nonconforming tiles in the sample. a) If 5% of the tiles produced are nonconforming, what is P(X ≥ 75)? b) Based on the answer to part (a), if 5% of the tiles are nonconforming, is 75 nonconforming tiles out of 1000 an unusually large number? c) If 75 of the sample tiles were nonconforming, would it be plausible that the goal had been reached? Explain. d) If 5% of the tiles produced are nonconforming, what is P(X ≥ 53)? e) Based on the answer to part (d), if 5% of the tiles are nonconforming, is 53 nonconforming tiles out of 1000 an unusually large number? f) If 53 of the sample tiles were nonconforming, would it be plausible that the goal had been reached? Explain.
An investigator analyzed the leading digits from 794 checks issued by seven suspect companies. The frequencies were found to be 218, 130, 106, 83, 65, 48, 48, 37, and 59, and those digits correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. If the observed frequencies are substantially different from the frequencies expected with Benford's law shown below, the check amounts appear to result from fraud. Use a 0.025 significance level to test for goodness-of-fit with Benford's law. Does it appear that the checks are the result of fraud? Leading_Digit Actual_Frequency Benford's_Law:1) 218 30.10%2 )130 17.60%3) 106 12.50%4) 83 9.70%5) 65 7.90%6) 48 6.70%7) 48 5.80%8) 37 5.10%9) 59 4.60% Calculate the test statistic, χ2 =? (Round to three decimal places as needed.) Calculate the P-value=? (Round to four decimal places as needed.)
I need help with a b and c of question 1 please! Image is attached.
3. The paper "Effects of Caffeine on Repeated Sprint Ability, Reactive Agility Time, Sleep and Next Day Performance"† describes an experiment in which male athlete volunteers who were considered low caffeine consumers were assigned at random to one of two experimental groups. Those assigned to the caffeine group drank a beverage which contained caffeine 1 hour before an exercise session. Those in the no-caffeine group drank a beverage that did not contain caffeine 1 hour before an exercise session. That night, participants wore a device that measures sleep activity. The researchers reported that there was no significant difference in mean sleep duration for the two experimental groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means. A) The difference in mean sleep duration for the caffeine and no-caffeine groups could have occurred by chance just due to the random assignment, assuming there is a real difference…