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S1 Mixed Starters

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Height (mm)Frequency 11-2 16-8 21-15 26-23 31-357 a)Calculate an estimate of the mean b)Calculate the standard deviation of the data

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Height (mm) f 11-2 16-8 21-15 26-23 31-357 Xfx 1326 18144 23345 28644 33231 Σf = 55Σfx = 1390 x²x²fx² 169338 3242592 5297935 78418032 10897623 Σfx² = 76520 a)x = Σfx = 1390 = 25.3 Σf 55 b) Variance = Σfx² - Σfx ² = 76520 - 1390 ² = 752.56 Σf Σf 55 55 Standard deviation = 752.56 = 27.43

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Venn diagrams and conditional probability P(C) = 0.4 P(D) = 0.7 P(C D) = 0.2 Calculate a) P(D C) P(C` D`) P (C` D)

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Calculate a) P(D C) = 0.14 ÷ 0.4 = 0.35 P(C` D`) = 0.04 P (C` D) = 0.56 P(C D) = P(C D) x P(D) P(C D) = 0.2 x 0.7 = 0.14 CD 0.140.260.56 S 0.04

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Weight (kg)Frequency 10 - 1420 15 - 1914 20 - 244 25 - 291 30 - 341 a) Perform the calculations that you would need to be able to draw a histogram (Do not draw the histogram) b) Work out Q1, Q2 and Q3 and comment on the skewness of the data set c) Calculate D 3 and P 67

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Weight (kg)Frequency 10 - 1420 15 - 1914 20 - 244 25 - 291 30 - 341 Weight (kg)Frequency density 9.5 – 14.520÷5=4 14.5 – 19.514 ÷5=2.8 19.5 – 24.54 ÷5=0.8 24.5 – 29.51 ÷5=0.2 29.5 – 34.51 ÷5=0.2 CF 20 34 38 39 40 Q1 = 12 Q2 = 14.5 Q3 = 18.1 Q3-Q2 = 3.6 Q2-Q1 = 2.5 Q3-Q2 > Q2-Q1 therefore there is positive skew D3 = 12.5 P67 = 17.9

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The probability that is snows on a particular day in winter is 0.24. If it snows the probability that I go to work is 0.18. If it does not snow the probability that I go to work is 0.85. a)Calculate the probability that it snows but I do not go to work b)Calculate the probability that I go to work

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The probability that is snows on a particular day in winter is 0.24. If it snows the probability that I go to work is 0.18. If it does not snow the probability that I go to work is 0.85. a)Calculate the probability that it snows but I do not go to work b)Calculate the probability that I go to work S S` W W` W 0.24 0.76 0.18 0.82 0.85 0.15 a) 0.24 x 0.82 = 0.1968 a) (0.24 x 0.18) + (0.76 x 0.85) = 0.6892

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Adam and Beth decide to play a game. The probability that Adam wins the game is 0.2 and the probability that Beth wins is 0.35. They decide to play the game 3 times. Given that the results of the game are independents, find a)P(wins all 3 games) b)P(all 3 games end in a draw) c)P(Adam wins 1 and Beth wins 2 games) d)P(Adam and Beth win one game each)

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Adam and Beth decide to play a game. The probability that Adam wins the game is 0.2 and the probability that Beth wins is 0.35. They decide to play the game 3 times. Given that the results of the game are independents, find a)P(Adam wins all 3 games) b)P(all 3 games end in a draw) c)P(Adam wins 1 and Beth wins 2 games) d)P(Adam and Beth win one game each) A B D A B D A B D A B D A B D A B D A B D A B D A B D A B D A B D A B D A B D a) 0.2 x 0.2 x 0.2 = 0.008 b) 0.45 x 0.45 x 0.45 = 0.091125 c) P(ABB)+P(BAB)+P(BBA) (0.2 x 0.35 x 0.35) x 3 0.0245 x 3 = 0.0735 d) P(ABD)+P(ADB)+P(BAD)+P(BDA)+P (DAB)+P(DBA) (0.2 x 0.35 x 0.45) x 6 0.0315 x 6 = 0.189

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In a college there are 100 students taking A level French, German or Spanish. Of these students, 64 are female and the rest are male. There are 50 French students of whom 40 are female and 30 German students of whom 10 are female. Find the probability that a randomly chosen student a)Is taking Spanish b)Is male, given that the student is taking Spanish College records indicate that 70% of the French students, 80% of the German students and 60% of the Spanish students have applied for university. A student is chosen at random c) Find the probability that this student has applied for University d) Given the student had applied for University, find the probability that the student is studying French

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FemaleMaleTotal French401050 German102030 Spanish14620 Totals6436100 a) P(Spanish) = 20 / 100 = 1 / 5 b) P(M|Spanish) = P(M S) = 6 / 100 = 6/ 20 = 3/ 10 S 20 / 100 Applied to University 70% of 50 = 35 80% of 30 = 24 60% of 20 = 12 Total = 71 c) P(Applied to Uni) = 71 / 100 d) P(French|Uni) = P(F U) = 35 / 100 = 35/ 71 U 71 / 100

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A researcher thinks there is a link between a person's height and level of confidence. She measured the height h, to the nearest cm, of a random sample of 9 people. She also devised a test to measure the level of confidence c of each person. The data are shown in the table below H179169187166162193161177168 C569561579561540598542565573 [You may use h 2 = 272 094, c 2 = 2 878 966, hc = 884 484] (a)Find exact values of S hc S hh and S cc. (4) (c)Calculate the value of the product moment correlation coefficient for these data. (3) (d)Give an interpretation of your correlation coefficient. (1)

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(b)S hc = 884484 – 1562x5088 = 1433 9 S hh = 1000 2 / 9 ; S cc = 2550 (c)r = 1433 1000 2 / 9 x 2550 = 0.897488…. (d)Taller people tend to be more confident

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The number of caravans on Seaview caravan site on each night in August last year is summarised in the following stem and leaf diagram. Caravans 1 0 means 10 Totals 10 5 (2) 21 2 4 8 (4) 30 3 3 3 4 7 8 8 (8) 41 1 3 5 8 8 8 9 9 (9) 52 3 6 6 7 (5) 62 3 4 (3) (a)Find the three quartiles of these data. (3) During the same month, the least number of caravans on Northcliffe caravan site was 31. The maximum number of caravans on this site on any night that month was 72. The three quartiles for this site were 38, 45 and 52 respectively. (b) On graph paper and using the same scale, draw box plots to represent the data for both caravan sites. You may assume that there are no outliers. (6) (c) Compare and contrast these two box plots. (3) (d) Give an interpretation to the upper quartiles of these two distributions.(2)

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a) Q 1 = 33, Q 2 = 41, Q 3 = 52 B1B1B13 b) Median of Northcliffe is greater than median of Seaview Upper quartiles are the same IQR of Northcliffe is less than IQR of Seaview Northcliffe positive skew, Seaview negative skew Northcliffe symmetrical, Seaview positive skew (quartiles) Range of Seaview greater than range of Northcliffe B1 B1 B1 3

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d) On 75% of the nights that month B1 both had no more than 52 caravans on site. B12

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A discrete random variable has the following probability distribution X123456 P(X=x)0.20.150.20.30.10.05 Find a)P(1 < X < 5) b)P(2 X 4) c)P(3 < X 6) d)P(X < 3)

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SOLUTION A discrete random variable has the following probability distribution X123456 P(X=x)0.20.150.20.30.10.05 P(1 < X < 5) = P(X=2,3 or 4) = 0.15 + 0.2 + 0.3 = 0.65 P(2 X 4) = P(X=2,3 or 4) = 0.15 + 0.2 + 0.3 = 0.65 P(3 < X 6) = P(X=4,5 or 6) = 0.3 + 0.1 + 0.05 = 0.45 P(X < 3) = P(X=1 or 2) = 0.2 + 0.15 = 0.35

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Eight students took tests in mathematics and physics. The marks for each student are given in the table below where m represents the mathematics mark and p the physics mark. Students ABCDEFGH MarkM9141310782017 p1123211519103126 A science teacher believes that students marks in physics depend upon their mathematical ability. The teacher decides to investigate this relationship using the test marks. (a)Draw a scatter diagram to illustrate these data. (3) (b)Showing your working, find the equation of the regression line of p on m. (8) (c)Draw the regression line on your scatter diagram. (2) A ninth student was absent for the physics test, but she sat the mathematics test and scored 15. (d)Using this model, estimate the mark she would have scored in the physics test. (2)

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(a)scales and labels B1 points B2 (6,7 points) B1 Line M1 A1 m = 98; p = 156; m 2 = 1348; mp = 2119 S mp = 2119 98 x 156 = 208 M1 A1 8 S mm = 1348 98 2 = 147.5 A1 8 b = Smp = 208 = 1.410169 (awrt 1.41) M1 A1 Smm 147.5 a = 156 (1.410169…) × 98 = 2.225429 (awrt 2.23)M1 A1 8 8 p = 2.23 + 1.41m A1 ft (c)Line on graph M1 A1 (d)p = 2.23 + 1.41 × 15 = 23.38 M1 A1

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