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# In a 50 marks non minus exam what's the probability of a student's gettin 30 or 35 marks ?

What's the probability of a student's getting 30 or 35 marks in a 50 marks exam where no minus marking is done.

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### Practice Questions Answers for Second Exam – 2012

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Practice Exam Questions ... 10, 20, 30, 40, 50, 60, 70, 80 ... amount of student loans at the present time (in ...

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Jun 04, 2015 · How to solve Hannah's Sweets: a former GCSE maths teacher ... missing out one of your marks for the ... have Hannah's sweets on their exam ...

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A diagram showing a sample space S and events as areas within S. Overlaps indicate non ... Probability: .05 .10 .35 .50... E(X ... a standard deviation of 30\$, ...

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... Student Guide | Page 1 Summary of Video ... Student Guide | Page 2 1 50 ... Distribution of the sample mean from 1000 samples of size 50. 16 18 20 22 24 26 28 30

### Hannah's Sweets: the GCSE maths problem that had students ...

Jun 04, 2015 · An exam question on the Edexcel GCSE maths paper this week has become an internet ... Hannah's Sweets: ... 6/10 X 5/9 = 30/90 or 1/3 . or: 6/n * 5/n-1 = 1 ...

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Probability of multiple-choice answers in questions. ... in the overall average percentage of marks? ... at least 40% is 1 minus probability of getting less than ...

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D. Estimate whether the association is linear or non ... be predicted to score 50 on the midterm exam. C. a student who scored 2 points higher than ... 35. The value ...

## Suggested Questions And Answer :

### In a 50 marks non minus exam what's the probability of a student's gettin 30 or 35 marks ?

1. how dum are the students? 2. dont yuno modern skuels kant fael NE student??? that wood impli sumthun rong with skuel (guvt) kant let talk like that start

### There are 10 girls and 20 boys in a class. If half of the girls and half of the boys have blue eyes. Find the probability that one student chosen as moniter is either a girl or have blue eyes?

If a student is chosen at random we have 4 possibilities: Boy blue eyes (10) Boy non-blue eyes (10) Girl blue eyes (5) Girl non-blue eyes (5) The probabilities associated with these are: 10/30, 10/30, 5/30, 5/30; or 1/3, 1/3, 1/6, 1/6. There are 15 students with blue eyes and there are 10 girls. The probability of picking a girl is 10/30=1/3; the probability of picking a blue-eyed of either gender is 15/30=1/2. The probability of picking a blue-eyed girl is 5/30=1/6. So if it's one or the other, but not both, we add 1/3 and 1/2 and subtract 1/6 because the probability 1/2 already includes blue-eyed girls, so we don't want to account for them twice. 1/3+1/2-1/6=4/6=2/3.

### A survey group of 5 students are chosen at random from 25 elementary students and 13high school students. Determine the probability that at least 2 students chosen are elementary students.

There are 38 students in total. The probability, p, of choosing an elementary student=25/38, and the probability of not choosing an elementary student, 1-p,  is 13/38. This is a binomial situation expressed by (p+(1-p))^5 (=1, certainty) which expands to: p^5+5p^4(1-p)+10p^3(1-p)^2+10p^2(1-p)^3+5p(1-p)^4+(1-p)^5. Each term has a meaning: p^5 is the probability of selecting 5 elementary students; 5p^4(1-p) the probability of exactly 4 elementary and one high school student; etc. The probability of at least 2 elementary students is the sum of the probabilities of exactly 2, 3, 4 or 5 elementary students; or it is 1 minus (sum of the probabilities of all high school and exactly one elementary school student)=1-(1-p)^5-5p(1-p)^4=1-(13/38)^5-5*25/38(13/38)^4=0.95026 or 95.03% approx.

### how to calculate pass mark if 60% of students pass

If 60% of 100 students passed, then 60 students passed. Let's say the pass mark was 45%, then it means that 60 students scored 45% or more. If the pass mark was 67%, then 60 students scored 67% or more.  The number of students obtaining a pass is not directly related to the pass mark. The probability that a student  selected randomly obtained a pass is 60%, but it doesn't determine the pass mark itself.

### in a recent survey, the following data were obtained in response to the question

7 students out of 58 have no opinion, so probability=7/58=12.07%. Probability of being a freshman is 28/58 and probability of a random student answering no is 7/58, so the chances of one or the other is 28/58+7/58=35/58=60.34%. 24/58=41.38% is the probability of a sophomore favouring the issue. Since a non-freshman means a sophomore and not favouring the issue means no or no opinion, the probability is 6/58=10.34%.

### If N is 1200, Mean is 55, S.D. is 10, Find , 80% of students are to be promoted, what should be the marks for promotion

If X is the number of marks required to be included in the 80%, and the mean is 55 with SD=10, we have the z score=(X-55)/10. Any score below X puts the student in the excluded 20%. The value of z corresponding to 0.2 is approximately -0.84, so X-55=-8.4, making X=46.6 or 47 to the nearest whole mark. Put X=46.6; z score=(46.6-55)/10=-0.84. P(-0.84)=1-P(0.84)=1-0.2=0.8=80% according to normal distribution tables. P(-0.84) is the probability of the marks being below the "pass" mark of 47 so that 80% of students will have sufficient marks for promotion. I looked up the z score for 80% or 0.8 and discovered that it corresponded roughly to z=0.84, so the z score below which the marks would not qualify for promotion is z=-0.84, the lower 20%. That's where I got -0.84 from.

### How can you get a probability of |X - 10|>1.8? given X~N (10,4)

Mean=10, SD=4. X-10>1.8, X>11.8 or 10-X>1.8, X<8.2. Convert to Z inequality: |X-10|>1.8, |X-10|/4>0.45, |Z|>0.45 means Z>0.45 or Z<-0.45. Z=(11.8-10)/4=1.8/4=0.45. This Z score corresponds to probability of 0.6736 (67.36%). But this probability is for Z<0.45, i.e., X<11.8. So we have to work out 1-0.6736=0.3264 (32.64%). Z=(8.2-10)/4=-1.8/4=-0.45. This Z score corresponds to probability of 1-0.6736=0.3264 (32.64%). So for probability of X<8.2 OR X>11.8, we add the two probabilities together: 0.6528 (65.28%). The easiest way to visualise this is to draw the normal distribution curve roughly and mark the Z=0 point which divides the curve into two equal halves. To the left of zero mark Z=-0.45 and to the right Z=0.45. The area to the right of Z=0.45 is where X>11.8, and the area to the left of Z=-0.45 is where X<8.2. These areas are the same, so we find one area and double it. P(-Z)=1-P(Z) so if Z=-0.45, we use Z=0.45 and subtract from 1. Double this to get the area (probability) expressed in the inequality.

### how we get that -0.84 value

If X is the number of marks required to be included in the 80%, and the mean is 55 with SD=10, we have the z score=(X-55)/10. Any score below X puts the student in the excluded 20%. The value of z corresponding to 0.2 is approximately -0.84, so X-55=-8.4, making X=46.6 or 47 to the nearest whole mark. Put X=46.6; z score=(46.6-55)/10=-0.84. P(-0.84)=1-P(0.84)=1-0.2=0.8=80% according to normal distribution tables. P(-0.84) is the probability of the marks being below the "pass" mark of 47 so that 80% of students will have sufficient marks for promotion. I looked up the z score for 80% or 0.8 and discovered that it corresponded roughly to z=0.84, so the z score below which the marks would not qualify for promotion is z=-0.84, the lower 20%. That's where I got -0.84 from.

### 4 poems by Shakespeare, 5 poems by Coleridge, 2 poems by Tennyson,

20. The two box-and-whisker plots below show the scores on a math exam for two classes. What do the interquartile ranges tell you about the two classes? (1 point) (0 pts) Class A has more consistent scores. (1 pt) Class B has more consistent scores. (0 pts) Overall class A performed better than class B. (0 pts) The lowest score was in class B. 1 /1 point 21. Is the following data set qualitative or quantitative? the numbers of hours spent commuting to work by the employees of a company (1 point) (0 pts) qualitative (1 pt) quantitative 1 /1 point The item below has been reviewed and is scheduled to be updated. All students will receive full credit for any response to the following. 22. Identify the sampling method. You want to determine the number of text messages students at your school send in a month. You randomly ask students in each of your classes. (1 point) (1 pt) random (0 pts) systematic (0 pts) stratified (0 pts) none of these 1 /1 point 23. A mathematics journal has accepted 14 articles for publication. However, due to budgetary restraints only 7 articles can be published this month. How many specific ways can the journal editor assemble 7 of the 14 articles for publication? (1 point) (0 pts) 14 (0 pts) 3,432 (0 pts) 98 (1 pt) 17,297,280 1 /1 point 24. Your English teacher has decided to randomly assign poems for the class to read. The syllabus includes four poems by Shakespeare, five poems by Coleridge, two poems by Tennyson, and two poems by Lord Byron. What is the probability that you will be assigned a poem by Coleridge, and then a poem by Lord Byron? (1 point) (0 pts) (0 pts) (1 pt) (0 pts) 1 /1 point The final score is 24/24 (100%).

### whats are all the zeros of 4x^5+28x^4-21x^3-147x^2+27x+189

Divide the quintic by 4: x^5+7x^4-21x^3/4-147x^2/4+27x/4+189/4. The factors of 189 are 3*3*3*7. So we could have as factors 3, 3/2, 3/4, 7, 7/2, 7/4. These are tentative solutions for x without taking account of the sign. To get + from 5 factors we have five pluses, 3 pluses and 2 minuses, 1 plus and 4 minuses. But we have negative coefficients, so we can rule out 5 pluses. We need to trial positive and negative factors. It seems unlikely that +7 is a factor because we have 4*7^5 and 28*7^4 governing the function. 28*7^4/147*7^2=4*7^2/21 which is a factor of about 10, showing that even 147x^2 is too small to have any significant effect on 28x^4. It seems more probable that -7 will be more effective as the odd powers of -7 will counteract the even powers to some extent, so -7 is a possible solution. We also have 7/2 and 7/4 to consider, both positive and negative, as well as 3, 3/2, 3/4. Synthetic division is probably the easiest way to check factors. Let's use it to check for -7: -7 | 4..28 -21 -147 27 189 ......4 -28....0..147...0 -189 ......4....0 -21......0 27 |....0 So x+7 is a factor and we have a quartic: 4x^4-21x^2+27. However, this is effectively a quadratic, because there are no odd powers. We quickly realise that the factors of 27 we need are 3 and 9, and the factors of 4 are 4 and 1 because 4*3+1*9=21: (4x^2-9)(x^2-3)=(2x-3)(2x+3)(x-sqrt(3))(x+sqrt(3)). Therefore the solution is: x=-7, 3/2, -3/2, sqrt(3), -sqrt3).