MZB127T1 Past Paper Calculus and Ordinary Differential Equations - PDF

1 SECTION A: Calculus and Ordinary Differential Equations Answer all questions. QUESTION 1 (a) Consider the multivariable function f (x, y) = x3 + y 2 − 2y. (i) Find both first order partial derivatives of f....

1 SECTION A: Calculus and Ordinary Differential Equations Answer all questions. QUESTION 1 (a) Consider the multivariable function f (x, y) = x3 + y 2 − 2y. (i) Find both first order partial derivatives of f. (2 marks) 2 fx = 3x , fy = 2y − 2 (ii) Find the gradient of the function f and the directional derivative of f at the point x = 0, y = 0, in the direction u = i + j. (3 marks) 2 ∇f (x, y) = 3x i + (2y − 2)j ⇒ ∇f (0, 0) = −2j [1 for gradient evaluated at point. Ok if evaluated after dot product] 1 1 1 Normalise vector: û = √ (i + j) = √ i + √ j [1 for normalising vector] 12 + 12 2 2 Directional derivative: √   1 1 Du f (0, 0) = −2j · √ i + √ j = − 2 = −1.414 [1 for dot product, no need 2 2 for decimal answer] (iii) Show that this function has a critical point at x = 0, y = 1. (1 mark) Substitute into partial derivatives to show they are both zero: fx (0, 1) = 0, fy (0, 1) = 2 − 2 = 0 [1: can either substitute or solve for x and y] (b) Consider the following iterated integral: Z 1 Z 2 I= xy dy dx. 0 2x (i) Calculate the value of the integral I. (3 marks) Z 1  2 2 y I= x dx [1 for inner integral success] 0 2 2x Z 1 = 2x − 2x3 dx [1 for correct substitution of inner limits] 0 1 x4  2 1 = x − = = 0.5 [1. Decimal value not required] 2 0 2 (ii) Sketch the region S in the (x, y) plane that this integral is being calculated over. (1 mark) MZB127T1.241 cont/...  2 y 2 S 1x [1: have to both sketch y = 2x and identify correct side.] MZB127T1.241 cont/...  3 QUESTION 2 (a) Consider the following first order linear differential equation with initial condition: dy 1 + y = 1, y(0) = 1. dt t + 2 (i) Show that the integrating factor is I(t) = t + 2. (1 mark) Integrating factor Z  1 I(t) = exp dt = exp(ln(t + 2)) = t + 2. t+2 [1 for formula—also award on off chance student notes multiplying by t + 2 results in LHS being a total derivative] (ii) Find the solution of the differential equation that satisfies the given initial condition. Multiply by integrating factor and integrate: t2 /2 + 2t + C Z (t + 2)y = (t + 2) dt ⇒ y =. t+2 [1 for integration, 1 for rearranging for y. Subtract 1 if C is in the wrong place.] C Initial condition: y(0) = = 1 so C = 2. [1, no need to put back into solution]. 2 (3 marks) (b) Consider the following second order differential equation: d2 y dy 2 + 4 + 3y = 2e2t. dt dt (i) Find the general solution of the homogeneous version of the problem. (2 marks) 2 Characteristic equation λ + 4λ + 3 = 0 [1, any variable used (e.g. m, r) is acceptable.] Solution of characteristic equation is λ = −3, −1, so homogeneous solution yH = C1 e−3t + C2 e−t (ii) Use the method of undetermined coefficients to find a particular solution of the non- homogeneous equation. (2 marks) Let yP = Ae2t. Substitute into ODE: 4Ae2t + 8e2t + 3e2t = 2e2t [1 for substituting into ODE] 2 Collect terms: (4 + 8 + 3)A = 2, so A =. Thus 15 2 2t yP = e [1 for finding constant. Don’t need to put back into solution]. 15 (c) Suppose that when solving a homogeneous, second order differential equation, you find that the roots of the characteristic equation are λ = 1 ± 5i. Which of the following plots could represent a solution of the ODE? Describe two features of the plot that justify your decision. (2 marks) MZB127T1.241 cont/...  4 Plot 1 Plot 2 0.5 3 2 0 1 -0.5 0 0 1 2 3 4 5 0 1 2 3 4 5 t t Plot 3 Plot 4 6 5 4 2 0 0 -5 0 1 2 3 4 5 0 1 2 3 4 5 t t Plot 4 is correct, since it features oscillations (λ complex) with increasing amplitude (real part of λ > 0) MZB127T1.241 cont/...  5 QUESTION 3 (a) Consider the following system of differential equations: y1′ (t) = −y1 (t) − 2y2 (t), y2′ (t) = −12y1 (t) + y2 (t). (i) Write the system of differential equations in matrix-vector form, identifying the co- efficient matrix A. (1 mark)   −1 −2 y′ = y [1. As long as matrix identified, ok.] −12 1 (ii) Show that the eigenvalues of the matrix A in part (i) are λ1 = 5 and λ2 = −5. (2 marks) −1 − λ −2 Find characteristic equation: det(A − λI) = = (−1 − λ)(1 − λ) − −12 1 − λ 24 = λ2 − 25 [1 for characteristic eqn, even if not simplified] So λ1 = 5, λ2 = −5. (iii) Find the eigenvectors of the matrix A corresponding to the eigenvalues in part (ii), and thereby form the general solution of the system of ODEs. (5 marks) [Important: eigenvectors could differ by a constant multiple! Do not penalise if correct but different from   below]  −6 −2 v11 For λ1 = 5, = 0 −12 −4 v21   1 Thus v21 = −3v11 or v1 = −3    4 −2 v12 For λ2 = −5, = 0 −12 6 v22   1 Thus v21 = 2v22 so v2 = 2 The general solution is thus     1 5t 1 −5t y = C1 e + C2 e. [1 for forming general solution] −3 2 (b) Suppose that a general solution of a different system of ODEs is found to be     1 t 2 −t y(t) = C1 e + C2 e. 1 0 Find the values of C1 and C2 so that the initial condition   2 y(0) = 2 is also satisfied. (2 marks) [Note: good student may see immediately by inspection C2 = 0 and C1 = 2 obviously a solution. Full marks in that case!] MZB127T1.241 cont/...  6       2 1 2 Apply initial condition: = C1 + C2 [On off-chance student does this with 2 1 0 solution from part a, do not penalise if working right] Solve for C1 and C2 :      1 2 C1 2 = [1 for setting up equation for constants, either in matrix or com- 1 0 C2 2 ponent form]        C1 1 0 −2 2 2 = − = , so C1 = 2 and C2 = 0 [1 for correct values of C2 2 −1 1 2 0 constants: note substitution into general solution not required. Also ok if algebraic equations solved in different way.] MZB127T1.241 cont/...  7 SECTION B: Probability and Statistics Answer all questions. QUESTION 4 (a) Consider the following table, which classifies a random sample of 200 people and the type of vehicle they drive: Two-wheel drive Four-wheel drive Petrol car 90 10 Electric car 5 20 Hybrid car (i.e. petrol and electric) 72 3 Use the data in the table to estimate: (i) The probability that someone drives a car that is both electric and four-wheel drive. (1 mark) Pr(electric AND four-wheel drive) = 20/200 = 0.1 [1 mark for the correct number. If they write the correct probability definition Pr(electric AND four-wheel drive) but calculate the number incorrectly, they get 0.5 marks.] (ii) The probability that someone’s electric car is two-wheel drive. (1 mark) Pr(two-wheel drive | electric) = 5/25 = 0.2 [1 mark. Same marking scheme as (i).] (iii) The probability that someone’s four-wheel drive is a hybrid. (1 mark) Pr(hybrid | four-wheel drive) = 3/33 = 1/11 [1 mark. Same marking scheme as (i).] (b) A medical condition is present in 10% of the population, and a test indicates the condition is present in 12% of the population. The test is 90% reliable for a patient who has the medical condition. What is the probability that a patient actually has the condition when the test suggests they do? To answer this question, let A be the event that the condition is present, and let B be the event that the test indicates the condition is present. Find the following: (i) Pr(A), Pr(B) and Pr(B|A). (1 mark) Pr(A) = 0.1. Pr(B) = 0.12. Pr(B|A) = 0.9. [1 mark for correct matching of the notation to the information in the question. 0.5 mark if one of the three is obtained correctly.] (ii) Use your answers to (i) to find the probability Pr(A|B), which is the probability that a patient has the condition when the test suggests they do. (2 marks) Pr(A|B) = Pr(B|A)Pr(A)/Pr(B) [1 mark for use of Bayes’ theorem (or other ap- propriate statistical formulae).] Pr(A|B) = 0.9 × 0.1 / 0.12 [1 mark awarded even if they don’t simplify the maths.] = 0.09/0.12 = 9/12 = 3/4. MZB127T1.241 cont/...  8 (c) The lifetime x (in years) of a particular electronic component follows the probability den- sity function ( 1.5 − 0.5x, 1 ≤ x ≤ 3, f (x) = 0, otherwise. (i) Based on this probability density function, what is the probability that the electronic component will last for more than 3 years? Explain your answer. (1 mark) Because the p.d.f. is zero for x > 3 [0.5 mark awarded for correct explana- tion/mathematical derivation], the probability that the electronic component lasts for more than 3 years is zero [0.5 mark awarded for correct probability]. (ii) Show that the probability that the electronic component lasts for more than 2 years is 25%. (3 marks) Z 3 Pr(X > 2) = (1.5 − 0.5x) dx [1 mark for correct integral. Give at least 0.5 2 mark if student showed they knew an integral with lower limit of 2 was required, even if the integral is wrong.] 3 0.5x2  = 1.5x − [1 mark for correct integration] 2 2 = 1.5 × 3 − 0.25 × 32 − (1.5 × 2 − 0.25 × 22 ) [1 mark for correct substitution of limits into the integral. No marks lost for subsequent problems in simplification of the maths.] = 4.5 − 2.25 − (3 − 1) = 2.25 − 2 = 0.25 MZB127T1.241 cont/...  9 QUESTION 5 (a) For the following dataset, which contains 4 values: 16 16 4 16 (i) Calculate the sample mean. (1 mark) X xi X= n 16 + 16 + 4 + 16 = [1 mark awarded, even if they don’t state the formula or 4 simplify the maths. Give at least 0.5 mark if student states the correct formula.] 52 = = 13. 4 (ii) Without performing any calculations, write the formula that would be used to ob- tain the sample standard deviation s from the dataset. (1 mark) 1 X 2  s2 = x2i − nX [1 mark awarded for stating this formula, or an n−1 equivalent correct formula. Lose 0.5 mark if denominator of fraction has n in- stead of n − 1.] (iii) It can be shown that the sample standard deviation for this dataset is s = 6. What is the sample variance? (1 mark) 2 s = 6 so sample variance s = 36. [1 mark for correct number. Give at least 0.5 mark if they know that there is a squared relationship √ between standard deviation and variance (even if they do it the wrong way, e.g. 6).] (iv) Calculate a 95% confidence interval for the true mean. (When calculating your an- swer, you may round numbers to two decimal places where necessary.) (3 marks) √ a = tn−1,α/2 × s/ n [1 mark for correct formula. Award this mark if they skip the formula and go correctly straight to the next line. Note that if they answer the whole question using the z-distribution instead, and they do so correctly, give a maximum of 2 marks for the whole question.] = t3,0.025 × 6/2 [0.5 mark for correct substitutions.] = 3.182 × 3 (from tables) [0.5 mark for correct usage of the tables based on their stated t-value.] ≈ 3.2 × 3 ≈ 9.6 Thus the 95% confidence interval is 13±9.6, or from 3.4 to 22.6. [1 mark for any attempt at describing the confidence interval, after a calculated.] (b) A specific brand of resistor is expected to have a nominal resistance of 1000 Ohms (mean) ± 50 Ohms (standard deviation). MZB127T1.241 cont/...  10 (i) If the resistances of individual resistors from this brand follow a normal distribu- tion, calculate the probability that a randomly selected resistor will have a resistance greater than 1150 Ohms. (3 marks)   1150 − µ Pr(X > 1150) = Pr Z > [1 mark for correct formula.] σ   1150 − 1000 = Pr Z > [0.5 mark for correct substitutions.] 50 = Pr(Z > 3) [0.5 mark for correct simplification.] ≈ 0.001 (from tables) [1 mark for correct usage of the tables based on stated z value.] (ii) Consider now that 10 resistors from the brand have had their resistance measured. Suppose that for these 10 resistors, the sample mean obtained is exactly 1000 Ohms and the sample standard deviation obtained is exactly 50 Ohms. If you were to cal- culate the probability that a randomly selected resistor will have a resistance greater than 1150 Ohms, based only on this sample information, would the probability you calculate be the same as the probability you calculated in part (i)? Explain your an- swer. (1 mark) No. [0.5 mark for saying No.] The calculated probability would require a t-distribution since it is being obtained from a sample mean and sample standard deviation. [0.5 mark for mentioning t-distribution.] MZB127T1.241 cont/...  11 QUESTION 6 (a) A soft drinks manufacturer claims that 90% of their customers prefer “cola flavour” over other soft drink flavours. (i) The soft drinks manufacturer surveys 4 customers. What is the probability that only one of the four customers prefers cola flavour over other soft drink flavours? Calculate the probability Pr(X = 1) using the binomial distribution and show that it equals 0.0036.   (2 marks) 4 Pr(X = 1) = 0.91 (1 − 0.9)3 [1.5 marks as follows: 0.5 mark for correct 1 formula. 1 mark for implementing correct values of n, x and p in the formula: if they get at least one of these three values correct, award a minimum of 0.5 mark out of 1.] = 4 × 0.9 × 0.13 [0.5 mark for evaluating the combination expression correctly. No marks lost for subsequent problems in simplification of the maths.] = 3.6 × 0.001 = 0.0036 (ii) The soft drink manufacturer conducts a survey of 100 people, and 85 of them prefer cola flavour over other soft drink flavours. In this case, the Central Limit Theorem holds. Carry out an appropriate hypothesis test using the z-distribution to determine whether this survey suggests that the true fraction of customers preferring cola flavour is different from 90%. (6 marks) H0 : p = 0.9. H1 : p ̸= 0.9. [1 mark as follows: 0.5 mark for stating p=0.9. 0.5 marks for using a ̸= instead of ≤ or ≥ in the alternative hypothesis.] ! p̂ − p Pr(p̂) ≈ Pr Z < p [0.5 marks] p(1 − p)/n ! 0.85 − 0.9 = Pr Z < p [0.5 marks] 0.9(1 − 0.9)/100 ! −0.05 = Pr Z < p 0.09/100 ! −0.05 = Pr Z < p 9/10000   −0.05 = Pr Z < 3/100   −5 = Pr Z < 3 = Pr(Z < −1.666...) = Pr(Z > 1.666...) [1 mark for applying symmetry of the distribution] which is between 0.025 and 0.05 according to the tables. [1 mark for correct usage of the tables based on stated z value] Two-sided test, so p-value between 0.05 and 1. [1 mark for doubling the probabil- ity based on two-sided test] MZB127T1.241 cont/...  12 Slight evidence against H0. [1 mark for converting p-value to evidence category.] (There is slight evidence that the true proportion of customers that prefer cola flavour is different to 90%.) [No marks lost for missing this statement.] (b) A linear regression analysis was applied to a dataset containing nine paired data points (x, y) to identify if the two variables x and y are linearly related to each other. The outputs of this regression analysis are shown below. Based on the outputs of the regression analysis, what is the strength of the evidence for a linear relationship between the two variables (y and x)? Justify your answer. (2 marks) p-value for the gradient is 0.72. [1 mark for correct identification of p-value] Hence there is insufficient evidence of a linear relationship. [1 mark for converting p-value to evidence category. Do not award this mark if no p-value is mentioned.] END OF PAPER MZB127T1.241  (i) RESOURCE SHEET Exp/log rules: e0 = 1, ln(1) = 0, ea+x = ea ex , ln(ax) = ln a + ln x, ea ln x = xa Differentiation: d a d ax d d d 1 x = axa−1 , e = aeax , sin(ax) = a cos(ax), cos(ax) = −a sin(ax), ln(x) = dx dx dx dx dx x d df dg Chain rule: f (g(x)) = dx dg dx d Product rule: f (x)g(x) = f ′ (x)g(x) + f (x)g ′ (x) dx Integration: ( 1 a+1 a ̸= −1 Z Z a x dx = a+1 x +c , eax dx = 1 ax e + c, ln(x) + c a = −1 a Z Z 1 1 sin(ax) dx = − cos(ax) + c, cos(ax) dx = sin(ax) + c a a Z Z du Integration by substitution: f (u(x)) dx = f (u) du dx Z Z Integration by parts: f (x)g ′ (x) dx = f (x)g(x) − f ′ (x)g(x) dx       u1 v1 a b Matrices and vectors: For u = ,v = , and M = : u2 v2 c d q Norm (magnitude): ||u|| = u21 + u22 u · v = u1 v1 + u2 v2 Dot (inner) product:      a b u1 au1 + bu2 Matrix-vector product: Mu = = c d u2 cu1 + du2   −1 1 d −b Determinant and inverse: det(M ) = ad − bc, M = det(M ) −c a Multivariable Calculus:   ∂f ∂f f Gradient: ∇f (x, y) = i+ j= x ∂x ∂y fy Directional Derivative: Du f (x0 , y0 ) = ∇f (x0 , y0 ) · û (û a unit vector) Critical point: x0 , y0 such that fx (x0 , y0 ) = 0 and fy (x0 , y0 ) = 0, that is, ∇f (x0 , y0 ) = 0    H > 0, fxx > 0 local min.   H > 0, f < 0 local max. fxx fxy xx Hessian determinant test: H = det ⇒ fyx fyy H < 0  saddle  H=0  indeterminate First order ODEs: Z Z dy 1 Separation of variables: = f (t)g(y) ⇒ dy = f (t) dt dt g(y) Z 1 R First order linear: y ′ (t) + p(t)y(t) = q(t) ⇒ y(t) = q(t)I(t) dt, I(t) = e p(t) dt I(t) MZB127T1.241  (ii) Second order ODEs:  λ1 t λ2 t c1 e + c2 e λ real distinct y ′′ (t) + by ′ (t) + cy(t) = 0  Characteristic equation: , y = (c1 + c2 t)e λt λ repeated ⇒ λ2 + bλ + c = 0   αt e [c1 cos(βt) + c2 sin(βt)] λ = α ± iβ Undetermined coefficients: Right hand side f (t) Particular solution yP (t) Constant a0 A0 Linear a0 + a1 t A0 + A1 t y(t)′′ + by(t)′ + cy(t) = f (t) Polynomial a0 + a1 t +... + an tn A0 + A1 t +... + An t n Exponential cekt Aekt Trigonometric a0 cos(kt) + a1 sin(kt) A0 cos(kt) + A1 sin(kt) Systems of ODEs: For linear ODE system y ′ (t) = Ay(t) where y a vector, A a matrix: Eigenvalues λ such that det(A − λn I) = 0, eigenvectors vn such that (A − λn I)vn = 0 for n = 1, 2. General solution (λ distinct): y = c1 v1 eλ1 t + c2 v2 eλ2 t Probability rules: Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B), Pr(A) = 1 − Pr(A), Pr(A|B) = Pr(A ∩ B)/Pr(B) Probability distributions for continuous variables: Z x Cumulative density function: F (x) = f (u) du = Pr(X < x) −∞ Z b Probability density function: Pr(a < X < b) = f (x) dx = F (b) − F (a) a Mean, median and variance: (P x p(x) if X is discrete Mean: µ = E(X) = R x f (x) dx if X is continuous Pr(X < m) = Pr(X > m) = 0.5 Median: (P 2  2  (x − µ)2 p(x) if X is discrete Variance: σ = Var(X) = E (X − µ) = R 2 (x − µ) f (x) dx if X is continuous Relationship between variance and mean: Var(X) = E(X 2 ) − (E(X))2 Sample mean, sample median and sample variance: n X Sample mean: X= xi /n i=1   n+1 Sample median: m= th observation 2 n n ! 2 1 X
2 1 X 2 Sample variance: s = xi − X = x2i − nX n − 1 i=1 n−1 i=1 MZB127T1.241  (iii) Binomial distribution: X ∼ bin(n, p), E(X) = np, Var(X) = np(1 − p),     n x n n! Pr(X = x) = p (1 − p)(n−x) , = x x x!(n − x)! µx e−µ Poisson distribution: X ∼ P(µ), E(X) = µ = λt, Var(X) = µ, Pr(X = x) = x! Exponential distribution: T ∼ Exp(λ), E(T ) = 1/λ, Var(T ) = 1/λ2 , f (t) = λe−λt , F (t) = 1 − e−λt , t≥0 Normal distribution: X ∼ N(µ, σ 2 ), Pr(X > x) = Pr(X < −x) X −µ x−µ Standard normal distribution: Z= ∼ N(0, 1), z= , σ σ Pr(Z > z) = Pr(Z < −z), Pr(Z > zp ) = p Student’s t distribution: T ∼ td , t∞ = z, Pr(T > t) = Pr(T < −t), Pr(T > td,p ) = p Central Limit Theorem: X is nearly normal when n ≥ 30, so for large (n ≥ 30) n 2 X −µ X −µ X ∼ N(µ, σX ), Z= = √ , σX σ/ n X −µ For (n < 30) T = √ , T ∼ tn−1 s/ n X ∼ bin(n, p) is nearly normal when n ≥ 30, np ≥ 5, and n(1 − p) ≥ 5, so p − p̂ X ∼ N(np, np(1 − p)), Z=p , p̂ = X/n p(1 − p)/n Confidence intervals: Confidence interval for a mean µ: X ± tn−1,α/2 √sn q Confidence interval for a proportion p: p̂ ± zα/2 p̂(1− n p̂) Hypothesis testing: Evidence against the null hypothesis p-value Very strong p < 0.01 Strong p ≈ 0.02 Moderate p ≈ 0.05 Slight 0.05 < p < 0.1 Insufficient p > 0.1 Linear regression: yi = β0 + β1 xi + εi , εi ∼ N(0, σ 2 ) Confidence interval in intercept: β̂0 ± tn−2,α/2 sβ̂0 Confidence interval in slope: β̂1 ± tn−2,α/2 sβ̂1 MZB127T1.241  (iv) Table of standard Normal percentage points Table of values zα such that Pr(Z > zα ) = α, for given α. Pr(Z > zα ) 0.5 0.25 0.2 0.15 0.1 0.05 0.025 0.01 0.005 0.001 0.0005 z0.5 z0.25 z0.2 z0.15 z0.1 z0.05 z0.025 z0.01 z0.005 z0.001 z0.0005 zα 0.000 0.674 0.842 1.036 1.282 1.645 1.960 2.326 2.576 3.090 3.291 Table of t–values Table of values tp such that Pr(t > tp ) = p, for given p and degrees of freedom (d.f.). Pr(t > tp ) 0.5 0.25 0.2 0.15 0.1 0.05 0.025 0.01 0.005 0.0005 d.f. ↓ t0.5 t0.25 t0.2 t0.15 t0.1 t0.05 t0.025 t0.01 t0.005 t0.0005 1 0.000 1.000 1.376 1.963 3.078 6.314 12.706 31.821 63.657 636.619 2 0.000 0.816 1.061 1.386 1.886 2.920 4.303 6.965 9.925 31.599 3 0.000 0.765 0.978 1.250 1.638 2.353 3.182 4.541 5.841 12.924 4 0.000 0.741 0.941 1.190 1.533 2.132 2.776 3.747 4.604 8.610 5 0.000 0.727 0.920 1.156 1.476 2.015 2.571 3.365 4.032 6.869 6 0.000 0.718 0.906 1.134 1.440 1.943 2.447 3.143 3.707 5.959 7 0.000 0.711 0.896 1.119 1.415 1.895 2.365 2.998 3.499 5.408 8 0.000 0.706 0.889 1.108 1.397 1.860 2.306 2.896 3.355 5.041 9 0.000 0.703 0.883 1.100 1.383 1.833 2.262 2.821 3.250 4.781 10 0.000 0.700 0.879 1.093 1.372 1.812 2.228 2.764 3.169 4.587 11 0.000 0.697 0.876 1.088 1.363 1.796 2.201 2.718 3.106 4.437 12 0.000 0.695 0.873 1.083 1.356 1.782 2.179 2.681 3.055 4.318 13 0.000 0.694 0.870 1.079 1.350 1.771 2.160 2.650 3.012 4.221 14 0.000 0.692 0.868 1.076 1.345 1.761 2.145 2.624 2.977 4.140 15 0.000 0.691 0.866 1.074 1.341 1.753 2.131 2.602 2.947 4.073 16 0.000 0.690 0.865 1.071 1.337 1.746 2.120 2.583 2.921 4.015 17 0.000 0.689 0.863 1.069 1.333 1.740 2.110 2.567 2.898 3.965 18 0.000 0.688 0.862 1.067 1.330 1.734 2.101 2.552 2.878 3.922 19 0.000 0.688 0.861 1.066 1.328 1.729 2.093 2.539 2.861 3.883 20 0.000 0.687 0.860 1.064 1.325 1.725 2.086 2.528 2.845 3.850 21 0.000 0.686 0.859 1.063 1.323 1.721 2.080 2.518 2.831 3.819 22 0.000 0.686 0.858 1.061 1.321 1.717 2.074 2.508 2.819 3.792 23 0.000 0.685 0.858 1.060 1.319 1.714 2.069 2.500 2.807 3.768 24 0.000 0.685 0.857 1.059 1.318 1.711 2.064 2.492 2.797 3.745 25 0.000 0.684 0.856 1.058 1.316 1.708 2.060 2.485 2.787 3.725 26 0.000 0.684 0.856 1.058 1.315 1.706 2.056 2.479 2.779 3.707 27 0.000 0.684 0.855 1.057 1.314 1.703 2.052 2.473 2.771 3.690 28 0.000 0.683 0.855 1.056 1.313 1.701 2.048 2.467 2.763 3.674 29 0.000 0.683 0.854 1.055 1.311 1.699 2.045 2.462 2.756 3.659 30 0.000 0.683 0.854 1.055 1.310 1.697 2.042 2.457 2.750 3.646 40 0.000 0.681 0.851 1.050 1.303 1.684 2.021 2.423 2.704 3.551 60 0.000 0.679 0.848 1.045 1.296 1.671 2.000 2.390 2.660 3.460 80 0.000 0.678 0.846 1.043 1.292 1.664 1.990 2.374 2.639 3.416 100 0.000 0.677 0.845 1.042 1.290 1.660 1.984 2.364 2.626 3.390 1000 0.000 0.675 0.842 1.037 1.282 1.646 1.962 2.330 2.581 3.300 ∞ 0.000 0.674 0.842 1.036 1.282 1.645 1.960 2.326 2.576 3.291 MZB127T1.241