Cambridge A Level Mathematics - Further 9231 — 2019 May/June Paper 2 · Variant 3

1 A bullet of mass 0.2 kg is fired into a fixed vertical barrier. It enters the barrier horizontally with speed 250 m s−1 and emerges horizontally after a time T seconds with speed 40 m s−1. There is a constant horizontal resisting force of magnitude 1200 N. Find T. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 M1 A1 Find impulse from change in momtm. (may be implied) (if sign of 40 wrong here or below, can allow M1) T = 42 / 1200 = 7/200 or 0⋅035 [s] M1 A1 Find time T from impulse = Ft Alternative method for question 1 a = 1200 / 0⋅2 = 6000 M1 A1 Find acceleration a from F = ma (may be implied) T = (250 – 40) / 6000 = 7/200 or 0⋅035 [s] M1 A1 Find time T from v = u + at Alternative method for question 1 K.E. = ½ × 0⋅2 × (2502 – 402) = 6090 M1 Find loss of K.E. (may be implied; ignore sign of K.E.) Work = 1200 × ½ (250 + 40) T = 174 000 T M1 A1 Find work done from F × ½ (u + v) t (A1 for both) T = 6090 / 174 000 = 7/200 or 0⋅035 [s] A1 Equate to find T 4

More questions on Momentum

2 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is moving in a complete vertical circle about O. The points A and B are on the circle, at opposite ends of a diameter, and such that OA makes an acute angle ! with the upward vertical through O. The speed of P as it passes through A is 3 ag . 2 The tension in the string when P is at B is four times the tension in the string when P is at A. (i) Show that cos ! = 34. 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(ii) Find the tension in the string when P is at B. 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Mark scheme: 2(i) B1 Find tension TA at A from F = ma radially TB = m vB2 / a + mg cos α B1 Find tension TB at B from F = ma radially ½mvB2 = ½m vA2 + 2mga cos α [vB2 = ag (9/4 + 4 cos α)] M1 A1 Apply conservation of energy at B (A0 if no m) vB2 = 4vA2 – 5ag cos α = vA2 + 4ga cos α M1 Combine using TB = 4 TA and vA2 = 9ag/4 to verify cos α 9ag cos α = 3vA2 = 27ag/4, cos α = 3/4 AG A1 6 2(ii) TB = 4 TA = (9 – 4 cos α) mg = 6mg or vB2 = 21ag/4, TB = (21/4 + cos α) mg = 6mg M1 A1 Find TB 2

More questions on Circular motion

3 Three uniform small spheres A, B and C have equal radii and masses 3m, m and m respectively. The spheres are at rest in a straight line on a smooth horizontal surface, with B between A and C. The coefficient of restitution between each pair of spheres is e. Sphere A is projected directly towards B with speed u. (i) Find, in terms of u and e, expressions for the speeds of A, B and C after the first two collisions. 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(ii) Given that A and C are moving with equal speeds after these two collisions, find the value of e. 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Mark scheme: 3(i) 3mvA + mvB = 3mu [or 3vA + vB = 3u] (AEF) M1 Use conservation of momentum for A and B (correct masses) vB – vA = eu M1 Use Newton’s restitution law with consistent LHS signs vA = ¼ (3 – e) u [ vB = ¾ (1 + e) u ] A1 Combine to find speed of A wB + vC = vB and vC – wB = evB (AEF) M1 Use conservation of momentum for B and C and Newton’s restitution law with consistent LHS signs wB = ½ (1 – e) vB and vC = ½ (1 + e) vB wB = (3/8) (1 – e2) u and vC = (3/8) (1 + e)2 u aef A1 A1 Combine to find wB and vC in terms of vB (may be implied) and with vB replaced (note: vC = 2 3 3 3 8 4 8 + + e u eu u ) 6 3(ii) ¼ (3 – e) = (3/8) (1 + e)2 M1 Find equation in e by equating their vA and their vC 3e 2 + 8e – 3 = 0 = (3e – 1) (e + 3) M1 Simplify and solve resulting quadratic eqn for e, e = 1/3 A1 (implicitly) rejecting e = –3 3

More questions on Momentum

4 A L a 7a 2a B An object consists of two hollow spheres which touch each other, together with a thin uniform rod AB. The rod passes through small holes in the surfaces of the spheres. The rod is fixed to the spheres so that it passes through the centre of the smaller sphere. The end B of the rod is at the centre of the larger sphere. The larger sphere has radius 2a and mass M, the smaller sphere has radius a and mass kM, and the rod has length 7a and mass 5M. A fixed horizontal axis L passes through A and is perpendicular to AB (see diagram). (i) Find the moment of inertia of the object, consisting of the rod and two spheres, about L. 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The object is pivoted at A so that it can rotate freely about L. The object is released from rest with the rod making an angle of 60Å to the downward vertical. The greatest angular speed attained by the 9 O@g Aobject in the subsequent motion is . 20 a (ii) Find the value of k. 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Mark scheme: 4(i) or (4/3) 5M (7a/2)2 [= (245/3) Ma 2] B1 Find or state MI of rod AB about axis at A IM = ⅔ M (2a)2 + M (7a)2 [= (155/3) Ma 2] M1 A1 M1 for one term correct, A1 for both terms correct IkM = ⅔ kM a 2 + kM (4a)2 [= (50/3) kMa 2] M1 A1 M1 for one term correct, A1 for both terms correct I = [ (245/3 + 155/3 + 50k/3) Ma 2 =] (50/3) (8 + k) Ma 2 A1 Find MI of object about axis at A, simplified to 2 terms aef 6 4(ii) ½ I ω2 = 5Mg (7a/2) (1 – cos 60°) + Mg (7a) (1 – cos 60°) *M1 A1 Find eqn. for ω 2 when AB vertical by energy (3 terms on RHS, same trig expression in each for M1) A1 for 2 terms correct on RHS + kMg (4a) (1 – cos 60° ) A1 A1 for RHS and LHS all correct ¼ (49 + 8k) / ½ (50/3) (8 + k) = 81/400 DM1 Equate ω2 to 81g / 400a to find k 4 (49 + 8k) = 27 (8 + k), k = 4 A1 Alternative method for question 4(ii) 6 + k) Mg × {(49 + 8k)/2(6 + k)} (1 – cos 60°) = ½ (49 + 8k) Mga (1 – cos 60°) = ¼ (49 + 8k) Mga *M1 A1 A1 M1 for mass × com × trig expression, A1 for 2 parts correct A1 all correct ¼ (49 + 8k) / ½ (50/3) (8 + k) = 81/400 DM1 Equate ω2 to 81g / 400a to find k 4 (49 + 8k) = 27 (8 + k), k = 4 A1 5

More questions on Equilibrium of a rigid body

5 A 1 a 2x B A uniform rod AB of length 2x and weight W rests on the smooth rim of a fixed hemispherical bowl of radius a. The end B of the rod is in contact with the rough inner surface of the bowl. The coefficient of friction between the rod and the bowl at B is 13. A particle of weight 4W1 is attached to the end A of the rod. The end B is about to slip upwards when AB is inclined at an angle 1 to the horizontal, where tan 1 = 3 (see diagram). 4 (i) By resolving parallel to the rod, show that the normal component of the reaction of the bowl on the rod at B is 4W.3 [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find, in terms of W, the reaction between the rod and the smooth rim of the bowl. 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(iii) Find x in terms of a. 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Mark scheme: 5(i) *M1 A1 FB = ⅓ RB B1 Relate FB and RB (may be implied) (4/5 + ⅓ × 3/5) RB = (1 + ¼) (3/5) W, RB = ¾ W AG DM1 A1 Combine using tan θ = ¾ to verify RA 5 5(ii) RC = FB cos θ – RB sin θ + W cos θ + ¼ W cos θ *M1 A1 Resolve forces perpendicular to rod AB, where C denotes rim Substitute to find RC = (⅓ × ¾ × 4/5) W – (¾ × 3/5) W + (5/4 × 4/5) W DM1 = (1/5 – 9/20 + 1) W = ¾ W A1 4 5(iii) A: RC × AC + (RB sin θ – FB cos θ ) × 2x – W cos θ × x [= W (¾ × 2x – ¾ × 8a/5 + 9/20 × 2x – 1/5 × 2x – 4x/5)] C: (RB sin θ – FB cos θ ) × 2a cos θ – W cos θ × CG + ¼W cos θ × AC [= W (9/20 × 2a – 1/5 × 2a – 8a/5 + x + ¼ × 2x – ¼ × 8a/5) cos θ ] G: RC × CG – (RB sin θ – FB cos θ ) × x – ¼ W cos θ × x [= W (¾ × 8a/5 – ¾ × x – 9/20 × x + 1/5 × x – 1/5 × x)] B: RC × 2a cos θ – W cos θ × x – ¼W cos θ × 2x [= W (¾ × 2a – x – ¼ × 2x) cos θ ] M1 A1 Find total moments about any point, denoting rod’s centre by G (taken on either side of C) where AC = 2x – 2a cos θ = 2x – 8a/5 CG = x – AC = 2a cos θ – x = 8a/5 – x x = a A1 Substitute and equate to zero to find x 3

More questions on Equilibrium of a rigid body

6 The random variable T is the lifetime, in hours, of a randomly chosen battery of a particular type. It is given that T has a negative exponential distribution with mean 400 hours. (i) Write down the probability density function of T. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that a battery of this type has a lifetime that is less than 500 hours. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the median of the distribution. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) [= 0 (t < 0)] B1 1 6(ii) P(T < 500) = ∫0500 f(t) dt = [exp (– t/400)] 0500 = 1 – e–5/4 = 0⋅7135 or 0⋅713 (allow 0⋅714) M1 A1 Find P(T < 500): (M0 for 1 – F(500) = 0.287) 2 6(iii) 1 – e-m/400 or e-m/400 = ½ , em/400 = 2 M1 Find median value m from F(m) = ½ m = 400 ln 2 or 277 M1 A1 3

More questions on Continuous random variables

7 A pair of fair coins is thrown repeatedly until a pair of tails is obtained. The number of throws taken is denoted by the random variable X. (i) State the expected value of X. [1] ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that exactly 3 throws are required to obtain a pair of tails. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that fewer than 4 throws are required to obtain a pair of tails. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Find the least integer N such that the probability of obtaining a pair of tails in fewer than N throws is more than 0.95. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) B1 Find or state E(X) 1 7(ii) P(X = 3) = q 2 p (with p = ¼, q = ¾) = 9/64 or 0⋅141 M1 A1 Find prob. of exactly 3 throws needed 2 7(iii) P(X < 4) = 1 – q 3 = 37/64 or 0⋅578 M1 A1 Find prob. of fewer than 4 throws needed 2 7(iv) 1 – q N – 1 > 0⋅95 (AEF) M1 Formulate condition for N (1 – q N is M0) 0⋅05 > (3/4) N – 1 , N – 1 > log 0⋅05 / log 0⋅75 M1 Set q = 3/4, rearrange and take logs (any base) to give bound N – 1 > 10⋅4, Nmin = 12 A1 Find Nmax (< or = can earn M1 M1 A0, max 2/3) 3

More questions on Continuous random variables

8 Two salesmen, A and B, work at a company that arranges different types of holidays: self-catering, hotel and cruise. The table shows, for a random sample of 150 holidays, the number of each type arranged by each salesman. Type of holiday Self-catering Hotel Cruise A 25 38 21 Salesman B 28 21 17 Test at the 10% significance level whether the type of holiday arranged is independent of the salesman. 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Mark scheme: 8 H0: Holidays are independent of salesman or no association between holidays and salesman (AEF) B1 Find expected values (lose A1 if rounded to integers) Find value of X 2 from Σ (Ei – Oi)2 / Ei [or Σ Oi2/Ei – n ] State or use correct tabular χ 2 value Ei: 29⋅68 33⋅04 21⋅28 (to 1 d.p.) 23⋅32 25⋅96 16⋅72 M1 A1 X2 = 0⋅7380 + 0⋅7446 + 0⋅0037 + 0⋅9392 + 0⋅9477 + 0⋅0047 M1 = 3⋅38 (to 3 s.f.) A1 χ2, 0.9 2 = 4⋅605 (to 3 s.f.) B1 Accept H0 if X 2 < tabular value M1 Compare their calculated value with their X 2 value and appropriate conclusion Type of holidays is independent of salesman (AEF) A1 Correct conclusion, from correct values ( ) 3.37 3.39 − 8

More questions on χ²-tests

9 A farmer grows large amounts of a certain crop. On average, the yield per plant has been 0.75 kg. The farmer has improved the soil in which the crop grows, and she claims that the yield per plant has increased. A random sample of 10 plants grown in the improved soil is chosen. The yields, x kg, are summarised as follows. Σx = 7.85 Σx2 = 6.19 (i) Test at the 5% significance level whether the farmer’s claim is justified, assuming a normal distribution. 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(ii) Find a 95% confidence interval for the population mean yield for plants grown in the new soil. 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Mark scheme: 9(i) B1 s 2 = (6⋅19 – 7⋅852 /10) / 9 [ = 37/12 000 or 0⋅003083 or 0⋅055532 ] M1 Estimate population variance (allow biased here: 0⋅002775 or 0⋅052682) H0: µ = 0⋅75, H1: µ > 0⋅75 (AEF) B1 State hypotheses (B0 for x …) t 9, 0.95 = 1⋅83[3] B1 State or use correct tabular t-value t = ( x – 0⋅75)/(s/√10) = 1⋅99 [Reject H0 and accept H1] M1 A1 Find value of t (or compare x with 0⋅75 + 0⋅032 = 0⋅782) Consistent conclusion Claim (of yield per plant increased) is justified (AEF) B1 FT on both t-values (must be t value) 7 9(ii) x ± t √(s 2 /10) M1 Find confidence interval (must be a t value) t 9, 0.975 = 2⋅26[2] B1 State or use correct tabular value of t 0⋅785 ± 0⋅04 or [0⋅745, 0⋅825] A1 Evaluate confidence interval (either form) 3

More questions on Inference using normal and t-distributions

10 The means and variances for a random sample of 8 pairs of values of x and y taken from a bivariate distribution are given in the following table. Mean Variance x 3.3125 3.3086 y 6.7375 7.9473 The product moment correlation coefficient for the sample is 0.5815, correct to 4 decimal places. (i) Find the equation of the regression line of y on x. 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(ii) Test at the 5% significance level whether there is evidence of positive correlation between x and y. 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(iii) Calculate an estimate of y when x = 6.0 and comment on the reliability of your estimate. 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Mark scheme: 10(i) b = Sxy / Sxx and r = Sxy / √(Sxx Syy) so M1 Relate gradient b in y = bx + c to r, Sxx , Syy b = r √( Syy / Sxx) A1 = 0⋅5815 √(7⋅9473 / 3⋅3086) = 0⋅901[2] M1 A1 Find Sxx and Syy and hence find b to 3 s.f. (y – 6⋅7375) = b (x – 3⋅3125), y = 0⋅901x + 3⋅75 M1 A1 Find equation of regression line of y on x Alternative method for question 10(i) Sxy = r √( Sxx Syy) = r √{(8sx2) (8sy2)} = 8 × 0⋅5815 √(7⋅9473 × 3⋅3086) M1 Find Sxy (allow values consistently scaled by factor 8) = 23⋅855 (allow 2⋅9818) (to 4 s.f.) A1 b = Sxy / Sxx = 23⋅85 / (8 × 3⋅3086) = 0⋅901[2] M1 A1 Hence find b to 3 s.f. (y – 6⋅7375) = b (x – 3⋅3125), y = 0⋅901x + 3⋅75 M1 A1 Find equation of regression line of y on x 6 Question Answer Marks Guidance 10(ii) H0: ρ = 0, H1: ρ > 0 B1 State both hypotheses (B0 for r …) r8, 5% = 0⋅621 B1 State or use correct tabular one-tail r-value Accept H0 if 0⋅5815 < tab. r-value (AEF) M1 State or imply valid method for conclusion No evidence of positive correlation (AEF) A1 Correct conclusion Alternative method for question 10(ii) H0: ρ = 0, H1: ρ > 0 B1 State both hypotheses (B0 for r …) tr = r√((n-2) / (1 – r 2)) = 1⋅75, t6,0⋅95 = 1⋅943 B1 Accept H0 if | tr | < tab. t-value (AEF) M1 State or imply valid method for conclusion No evidence of positive correlation (AEF) A1 Correct conclusion 4 10(iii) y = 9⋅16 B1 Find y when x = 6⋅0 Unreliable since r is small or r is not close to 1 or no correlation or 6⋅0 is not close to mean of x (AEF) B1 Reasonable comment on reliability 2 Question Answer Marks Guidance 11E(i) ½ mu 2 = ½ kmg (¼ a)2/a , k = 16u 2/ag AG M1 A1 Verify k by using conservation of energy (M0 if ω is found and SHM formula v = aω then used) 2 11E(ii) ± m d2x/dt 2 = kmg x/a *M1 Apply Newton’s law at general point (e.g. OP = a + x), requires m d2x/dt 2 = – (kg/a) x or – (16u 2 / a 2) x [ω = √ (kg/a) or 4u/a] A1 Derive standard SHM form (requires minus sign) Period is 2π√(a / kg) = 2π√(a 2 / 16u 2) = πa / 2u DM1 A1 Find period from 2π/ω 4 11E(iii) [½ mv 2 = ¾ × ½ mu 2] v 2 = ¾ u 2 or v = (√3/2) u M1 Relate v and u using given loss in energy x = (¼ a) sin ωt, v = (¼ a) ω cos ωt or u cos ωt M1 Relate v and t cos ωt = (√3/2) u / (¼ a) ω = (√3/2) M1 A1 Combine to find cos ωt t = (π/6) / (4u/a) = πa / 24u or 0⋅0417 πa / u or 0⋅131 a/u M1 A1 and hence t 6 Question Answer Marks Guidance 11O x – y = 0⋅448 – 0⋅48 = [–] 0⋅032 B1 Find difference in sample means (either sign; may be implied) sX 2 = (10⋅1 – 22⋅42 /50) / 49 = 0⋅0648 / 49 [= 0⋅001322] M1 A1 Estimate both population variances (may be implied) sY 2 = (16⋅3 – 28⋅82 /60) / 59 = 2⋅476 / 59 [= 0⋅04197] A1 (allow biased here: 0⋅001296 and 0⋅04127) sC 2 = sX 2/50 + sY 2/60 M1 Estimate combined variance = 0⋅0007259 or 0⋅026942 (to 3 s.f. throughout) A1 z = 0⋅032 / sC [or 0⋅032 / sP √(1/50 + 1/60)] = 1⋅188 [or 1⋅0895] M1 A1 Find value of z (either sign) Φ(z) = 0⋅8826 [or 0⋅8620] A1 Find Φ(z) 100 × (1 – Φ(z)) = 11⋅7 [or 13⋅8] M1 A1 Find limiting value for α, based on one-tail test (M0 for basing on two-tail test) α > (or ⩾) 11⋅7 [or 13⋅8] A1 Find set of possible values of α Allow 11.8 if 11.74 seen (Misreading α % as α loses only last A1) Question Answer Marks Guidance 11O Alternative method for question 11O x – y = 0⋅448 – 0⋅48 = [–] 0⋅032 B1 Find difference in sample means (either sign; may be implied) sX 2 = (10⋅1 – 22⋅42 /50) / 49 = 0⋅0648 / 49 [= 0⋅001322] M1 A1 Estimate both population variances (may be implied) sY 2 = (16⋅3 – 28⋅82 /60) / 59 = 2⋅476 / 59 [= 0⋅04197] A1 (allow biased here: 0⋅001296 and 0⋅04127) Assume equal [population] variances B1 State assumption sP2 = (49 sX 2 + 59 sY 2) / 108 or (0⋅0648 + 2⋅476) / 108 = 0⋅02353 or 0⋅15342 B1 (Find pooled estimate of common variance sX 2 and sY 2 not needed explicitly so may be implied by result) z = 0⋅032 / sC [or 0⋅032 / sP √(1/50 + 1/60)] = 1⋅188 [or 1⋅0895] M1 A1 Find value of z (either sign) Φ(z) = 0⋅8826 [or 0⋅8620] A1 Find Φ(z) 100 × (1 – Φ(z)) = 11⋅7 [or 13⋅8] M1 A1 Find limiting value for α, based on one-tail test (M0 for basing on two-tail test) α > (or ⩾) 11⋅7 [or 13⋅8] A1 Find set of possible values of α Allow 11.8 if 11.74 seen (Misreading α % as α loses only last A1) 12

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11 Answer only one of the following two alternatives. EITHER A light spring has natural length a and modulus of elasticity kmg. The spring lies on a smooth horizontal surface with one end attached to a fixed point O. A particle P of mass m is attached to the other end of the spring. The system is in equilibrium with OP = a. The particle is projected towards O with speed u and comes to instantaneous rest when OP = 34a. 16u2 (i) Use an energy method to show that k = . 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(ii) Show that P performs simple harmonic motion and find the period of this motion, giving your answer in terms of u and a. 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(iii) Find, in terms of u and a, the time that elapses before P first loses 25% of its initial kinetic energy. 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OR A company produces packets of sweets. Two different machines, A and B, are used to fill the packets. The manager decides to assess the performance of the two machines. He selects a random sample of

More questions on Hooke's law

50 packets filled by machine A and a random sample of 60 packets filled by machine B. The masses of sweets, x kg, in packets filled by machine A and the masses of sweets, y kg, in packets filled by machine B are summarised as follows. Σx = 22.4 Σx2 = 10.1 Σy = 28.8 Σy2 = 16.3 A test at the !% significance level provides evidence that the mean mass of sweets in packets filled by machine A is less than the mean mass of sweets in packets filled by machine B. Find the set of possible values of !. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................

More questions on Inference using normal and t-distributions