One question per chapter, 5 in all, taken from the course's corrected exercises. Answer the ones you can do and skip those from a chapter you have not seen yet: each answer is checked right away, and the solution opens after.
With your free account, each answer is kept and your dashboard files each chapter: passed the check, or to work on. These are the same questions as on the set pages: an answer given here counts there.
Two marks obtained in two different groups cannot be compared directly. The z-score brings them back to a common scale by measuring the distance from the mean in number of standard deviations.
Anne scores 78 in a group with a mean of 70 and a standard deviation of 8. Benoît scores 85 in a group with a mean of 80 and a standard deviation of 10.
The figure below shows the situation.
b) Which of the two has the better RELATIVE performance? Does the result surprise you?
See the solution
Answer
b)Anne, despite a lower raw mark: she is a full standard deviation above her group.
b) ANNE, with a z-score of 1.00 against 0.50. The result is surprising at first, since her raw mark, 78, is LOWER than Benoît's 85. But Anne stands a full standard deviation above her group, while Benoît is only half a standard deviation above his. Measured against the difficulty of the test and the spread of the group, Anne's performance is therefore better. This is exactly the reasoning behind the R score used for university admission in Quebec.
A carnival game: you bet 3 dollars, roll a fair die, and receive 12 dollars if a 6 comes up, 6 dollars if a 5 comes up, nothing otherwise. Let X be the player's NET GAIN, after deducting the bet.
a) Write down the probability distribution of X: the possible values and their probabilities. Check that the probabilities add up to 1.
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Answer
a)X=+9 with 61, X=+3 with 61, X=−3 with 64. Sum of the probabilities: 1.
a) Three possible outcomes. The 6 pays 12 dollars for 3 bet, so X=+9 with probability 61. The 5 pays 6 dollars, so X=+3 with probability 61. The four other faces pay nothing, so X=−3 with probability 64. Check: 61+61+64=1.
Chapter 3 · Continuous random variables and density
We keep the steel bar of exercise 2 and its density f(x)=8x on [0,4].
The CUMULATIVE DISTRIBUTION FUNCTION is the function F defined by F(x)=P(X≤x): it is the area accumulated under the density from the left up to the value x. It answers every probability question of the chapter in one go.
The figure below shows the situation.
a) Show, using the area of a triangle, that F(x)=16x2 for 0≤x≤4. Give F(2) and F(4), and the expression of F outside [0,4].
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Answer
a)F(x)=16x2 on [0,4], with F(2)=0.25 and F(4)=1; F=0 before 0 and F=1 after 4.
a) For x between 0 and 4, the accumulated area is that of the right triangle with base x and height f(x)=8x, namely F(x)=21×x×8x=16x2. No integration: a triangle is enough. We get F(2)=164=0.25 and F(4)=1616=1, which is the mandatory check: the cumulative distribution function must reach 1 at the end of the support. Outside, F(x)=0 for x<0, since no area has been accumulated yet, and F(x)=1 for x>4, since all of it has. Compare with part c) of exercise 2: F(2)=0.25 indeed gives P(X≤2) back.
A hypothesis test proves nothing: it measures how surprising the data would be if the default hypothesis were true, and compares that surprise with a threshold set IN ADVANCE.
A manufacturer states that its bags weigh 500 g. An inspector weighs 40 bags and finds a mean of 494 g. The standard deviation of the process is known, σ=15 g.
The figure below represents the situation.
c) Compute the p-value and state its exact meaning.
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Answer
c)p-value =2(0.0057)=0.0114: the probability of such a gap IF the bags really weighed 500 g.
c) Since the test is two-sided, the p-value is 2×P(Z<−2.530)=2×0.0057=0.0114. Exact meaning: IF the bags really weighed 500 g on average, then a sample of 40 bags differing from that value by at least 6 g, in either direction, would occur in only 1.14% of cases. The p-value is NOT the probability that H0 is true, a confusion that is the most widespread misinterpretation in statistics.
The square of the correlation coefficient is read directly as a percentage: the share of the variation in y that the model explains. It is the indicator quoted to judge the quality of a fit.
Keep r=0.9985 and the line y=20.77+6.67x, valid for x between 2 and 11 hours.
The graph below shows the data.
a) Compute the coefficient of determination and state its interpretation.
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Answer
a)r2=0.9971: 99.7% of the variation in the grades is explained by study time.
a) r2=0.99852=0.9971, that is, 99.7%. Interpretation: 99.7% of the observed variation in the grades is EXPLAINED by the linear relationship with the number of hours of study. This is an exceptionally good fit.
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