A teacher wants to grade a test on a curve, the cut-off scores would be established as follows:\ntop 10%=A; next 25%=B; next 30%=C; next 20%=D; last 15%=F\non normal curve, what are the cut-off scores. Mean is 74% and standard deviation is 15%\nfind all the cut-off scores
129547
Using the z-scores corresponding to the given areas, the cutoff scores would be:\nA: 93% (93.2233%)\nB: 80% (79.7798%)\nC: 68% (68.2202%)\nD: 58% (58.4535%)\nF: <58%
484282
Make a fraction putting the marks obtained by you as the numerator and the total marks as the denominator and then multiply the fraction by 100.you get your perentage.So your percentage is (308/405)X100=76%.So hard luck.You got C grade
635347
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
244937
Draw a scatter-plot. In short draw a graph of y against x.\nIf the graph show linear relationship, u may continue. If the graph is not linear(or curvilinear) ignore the questions as regression model is  only for linear model.\n\nSince u didn't include datas, i can only gives u equations\n\n\nQ1....\nthe mean for the number of years a teacher has taught \n    =  (∑ x)/n\n\n \nthe mean for the students' average course grade \n    =  (∑ y)/n\n\nQ2....\nthe variance for the number of years a teacher has taught \n     s²=[ ∑ (xi – mean for x value) ² ]/(n-1)\n\nthe variance for the students' average course grade\n     s²=[ ∑ (yi – mean for y value) ² ]/(n-1)\n\n\nQ3....\nthe standard deviation(s)\ns= √variance for the number of years a teacher has taught\n\ns= √variance for the students' average course grade\n\nQ4....\nthe z-score for the number of years a teacher has taught\n= (observed value of x-mean for x value)/ the standard deviation of x value\n\nthe z-score for the students' average course grade\n= (observed value of y-mean for y value)/ the standard deviation of y value.\n\nobserved value means u must calculate for each value one by one.\n\nQ5.....\nThe correlation coefficent(r)\n=1/(n-1) X  ∑(xi-mean value for x / s for x value)(yi- mean value for y)\n\ns stands for standard deviation and X is times and n is sample size\n\nThe equation looks quite tough isn't it? Normally we calculated r².\n\nr²= (SSTO-SSE)/SSTO\n\nwhere SSE = sum of squared errors\n          = ∑(yi-predicted value of y)²\n\nthe predicted value is base on the least square line( u have to draw a graph that has the sum of ( yi-predicted value of y) is smaller than any other line.) \n\n  \n     SSTO = sum of squares total\n          = ∑(yi-the mean for y value)²\n\nAfter u find r², find r by √r²\n      \nThe value of r always between -1 and +1\n#If u get value for r outside that range, means u have made some mistakes in calculation.\n#If u get r=1 or -1, it means that the relationship is perfectly linear and all data points fall on the same straight line.\n#r=+something means that when two varibles tend to increase together.(x increase, y increase)\n#r=0 means the best straight line is exactly horizontal\n\nQ6....\ni dont know about the coefficient of determination. Never heard such thing. Sorry.\n\nQ7....\nthe regression equation\n\ny-hat=b-nought  -  b-one   x\n\n  where    \nb-one = [∑(xi-the mean for x value)(yi-the mean for y value)]/  ∑(xi-the mean for x value)²\n\nb-nought = the mean for y value  -  b-one(the mean of x value)\n\nQ8....\n From y-hat=(b-nought)  -  (b-one) x\nJust simply substitude x=x1, therefore\n   predicted y = b-nought)  -  (b-one)x1\n\nQ9....\nI'm not sure about standard error of the estimate for predicting Y from X. But if if i'm not mistaken, the formula should be\n\nstandard error(s.e)= s/ [∑(yi-the mean for y value)²][b-one)\n\nwhere b-one=r(the standard deviation for x value)/(the standard deviation for y value)\n\nand s= √[SSE/(n-2)]\n\n\n\n\nGood luck
116117
If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
424928
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n,
353781
Although your question first appears overly brief, perhaps it does have enough information to give an "interval about 36.84" that encompasses only 10% of the bell curve's area.\n\nMy estimation is that we should therefore choose a value for alpha 'a' that is HALF the area you want ... because the "shaded area" will extend symmetrically on either side of the mean ... 5% each way.\n\nMean = 'mu' = u = 36.84\nStd Dev = 'theta' = o = 3.34\nProbability area = p = 0.50 - 0.05 = 0.45 (left side of curve)\nCritical value = z = 1.65 [from statistical table at back of textbook]\n\nWe wish to find the value of the "raw score" x that corresponds to the given z; we will simply use symmetry to find x2 from x1.  The solution will be given in interval notation [x1,x2], the "confidence limits."\n\nz = -1.65 = [(x - 36.84)/3.34]  NOTE: z is negative for "left side"\n\nSolving for x, we find x1 = 31.33 (rounded).\n\nRepeating the same calculation with positive z= 1.65, we find x2 = 42.35 (rounded).\n\nThe interval I believe you are seeking is [31.33,42.35].\n\nHope this helps!\n\nThe link below contains several Java-based statistical tools you may find helpful.\n\nHope this helps!
512749
Directions: Turn the following to percents.\n#1   10%\n\n#2   50%\n\n#3   60%\n\n#4   17%\n\n#5   75%\n\n#6   20%\n\n#7   30%\n\n#8   24%\n\n#9   12.5%\n\n#10   50%\n\n#11   225%\n\n#12   40%\n\n#13   25%\n\n#14   175%\n\n#15   40%\n\n#16   65%\n\n#17   .1%\n\n#18   125%
257550
95%CI = Mean ± (1.96 * (Sample Standard Deviation) / root(Sample Size))\n\n= 114 ± (1.96 * 13)/ root (50)\n=114 ± 3.6 = 110.4 to 117.6\n\nThe answer is (C)
212315
By definition, your score is above average.  IQ is a statistical measure, therefore your result is a function of how well you do relative to everyone else who has taken the test.  An IQ of 100 is the mean score.  15 points is equivalent to 1 standard deviation, therefore 68% of the population scores between 85 and 115.  If you plot your score on a normal curve (mean of 100, standard deviation of 15), the percentile score is 94.5%.  you scored better than 94% of the population.
674432
On the normal table, the z score such that 75% is to the left is 0.67.  That means that the 75th percentile is 0.67 SD's above the mean.  Since the normal is symmetric, the 25th percentile is 0.67 SD's below the mean.  So, to get the middle 50%, add and subtract .67*10 = 6.7 from 30.  The range is 23.3 and 36.7.  This is assuming that the population is normal.
495271
first fin the mean with the sample of 280 \n\nmultiply 280 to .184 to get the mean \n\nmultiply 280 to .184 and to .816 (from 1-.184) then take the \n\nsquare root that will be your standard deviation \n\n\ntake the z score you will get   ( 56 - 51.52 )/6.483851\n\n=.6909   find teh valu in the standard normal distribution table \n\nyou will get 0.2549    subtract it from .5  because you need only \n\nthe probability higher than of 56  you will get \n\n24.51%
481117
I assume a C = 70%\n\nSince there is a difference of 8 percentage points to 70.\nEight percentage points must be added to 70 to get the average.\n\n62 + 78 = 140\n\n140 is the total points needed to obtain a C\n\n140 divided by two is 70%\n\n- - - - - -s-
572395
You must  write your question more understandable.\n\nFor the test results, you can calculate the average, variance, standard deviation, min and max grades. You can also find the frequency of the marks.
564259
Grade Point Average...\n\n4.0=A\n3.0=B\n2.0=C\n1.0=D\n0.0=F\n\nIts just a number that shows colleges how well you do as a student.
542840
For question 1, think about the fraction. It's 9/24. A pie chart shows a fraction visually. Imagine a pie cut into 24 identical pieces. One piece for each student. Now color 9 of the pieces (all together) a different color from the rest of the pie to represent the newspaper only pupils.  Now measure the angle formed by the section of newspaper only). What is the angle of a sector of a circle that is 9/24 of the circle?\n\nFor question 2, a mixed number means there is a whole number and a fraction. What is .25?? It's 25 percent or one quarter of a dollar... so the fraction is 1/4. 2.25 is the same as 2 1/4... those digits add up to 7.
479059
It was my understanding that curves were given by teachers who thought that maybe they were teaching at a higher level than the children were supposed to achieve. I thought maybe they did this to be fair to children who were not at the same level as others. I did not know that colleges did this kind of thing. I would suggest that if it is a norm for the college to demand curves, start off every student with a negative score before the class begins. Maybe every ones average at a negative 15. Does this make any sense?