Normal distribution questions are found in paper 2, section B (Statistics section) of the H2 A Level Math exam. In this post, you will find the worked solutions for past A Level questions on normal distribution
A list of past questions from each statistics topic can be found here.
A quick summary of formulae in each statistics topics can be found here.
Past A Level Normal Distribution Questions
These are the past A Level normal distribution questions for the new syllabus (syllabus 9758) which started from 2017.
Click on the link to go straight to the worked solutions for each question.
Normal Distribution Question: 2017 Paper 2 Question 10
(i) Let the mass of the sphere be S grams.
S~N(20, 0.52)
P(S>20.2) = 0.34457 = 0.345 (3 s.f.)
(ii) Let the mass of the coated sphere be C grams.
C= 1.1S
C~ N(20 x 1.1, 1.12 x 0.52)
P(21.5 < C< 22.45) = 0.61172 = 0.612 ( 3 s.f.)
(iii) Let the mass of the metal bar be B grams.
B~N(µ, σ2 )
P(B>12.2) = 0.6
P(Z > (12.2- µ)/σ) = 0.6
(12.2- µ)/σ = – 0.2533471
µ – 0.2533471σ = 12.2 —(1)
P(B<12) = 0.25
P(Z > (12- µ)/σ) = 0.25
(12- µ)/σ = -0.6744897
µ – 0.6744897σ =12 —– (2)
Using graphic calculator,
mean = µ = 12.320 = 12.3 (3 s.f.)
standard deviation = σ = 0.47489 = 0.475 (3 s.f.)
(iv)
C1 +C2 + B ~ N(20 x 1.1 +20 x 1.1 + 12.320 , 1.12 x 0.52+ 1.12 x 0.52 + 0.474892) = N(56.320, 0.830520)
P(C1 +C2 + B > k ) = 0.75
k = 55. 705 = 55.7 ( 3 s.f.)
Normal Distribution Question: 2018 Paper 2 Question 10
(i) Let the mass of the light bulb be X.
X~ N(50, 1.52)
Using the graphic calculator, P(40<X<60) = 1
(ii) P(X<50.4) = 0.605137 = 0.605 (3 s.f.)
(iii) Let the mass of an empty box be B grams
B~N(75, 22)
Let T = B1 + B2 + B3 + B4
T~ N(300, 16)
P(T > 297) = 0.77337 = 0.773 ( 3 s.f.)
(iv) Let W = B + X
W~ N(125, 6.25)
P(124.9 < W < 125.7) = 0.126214 = 0.126 (3 s.f.)
(v) Let Y be the total mass of the bulb, padding and box in grams.
Y = 1.3X + B
Y~ N(1.3(50)+75, 1.32(1.52) + 22) = N(140, 7.8025)
P(Y>k) = 0.9
k = 136.420 = 136 ( 3.s.f)
(vi) Let C = Y1 + Y2 + Y3 + Y4
C~ N(140 x 4 , 7.8025 x 4) = N(560, 31.21)
P(C>565) = 0.185393 = 0.185 ( 3 s.f.)
Normal Distribution Question: 2019 Paper 2 Question 11
Let the mass of white balls be W grams. W~N(110, 42)
Let the mass of black balls be B grams. B~ N(55, 22)
(i) Let A = W1 + W2 + W3 + W4
A~N(440, 64)
P(A > 425) = 0.96960 = 0.970 ( 3 s.f.)
(ii) Let C = W + B
C~ N(165, 20)
P(161<C<175) = 0.801779 = 0.802 ( 3 s.f.)
(iii) Let T = B1 + B2 + B3 + W1 + W2
T~N(55 x 3 + 110 X 2, 4x 3 + 16x 2)
P(T<M) = 0.271
M = 380.955 = 381 ( 3.s.f)
(iv) Let the mass of the rod be R grams.
R~ N(20, 0.92)
Let mass of methane molecule be X grams.
X~ 0.7W + 0.9B1 + 0.9B2 + 0.9B3 + 0.9B4 + R1 + R2 + R3 + R4
X~ N(355, 24.04)
P(X>350) = 0.84608 = 0.846 ( 3.s.f)
Normal Distribution Question: 2020 Paper 2 Question 6
(i) T ~ N( 5, 1.22)
at 8 a.m., T = 0
at 8.10 a.m., T = 10
P(0<T<10) = 0.999969073 = 1.00 ( 3.s.f)
(ii) P ( T > 6) = 0.20232 = 0.202 ( 3 s.f.)
(iii) W~ N( 21, 32)
W + T ~N(26, 10.44)
P( James is late for work when he walks ) = P(W+T > 30) = 0.107863 = 0.108 ( 3 s.f.)
(iv)D~N( 19, 62)
D + T ~ N(24, 37.44)
P(James is late and weather is fine) = 0.7 (0.107863) = 0.0755041
P(Janes is late and weather is not fine) = 0.3 P(D+T > 30) = 0.0490199
P(James is late) = 0.0755041 + 0.0490199 = 0.124524
P( weather is fine| James is late)
= P(James is late and weather is fine) / P(James is late)
= 0.0755041/ 0.124524 = 0.60634 = 0.606 ( 3 s.f.)
Normal Distribution Question: 2021 Paper 2 Question 10
(a) Let the mass of the seat be S kg.
S~ N(µ, σ2 )
P(S<2.1) = 0.8
P(Z <(2.1 – µ)/σ) = 0.8
(2.1 – µ)/σ =0.841621
µ + 0.841521σ = 2.1 — (1)
P(S < 1.95) = 0.15
P(Z <(1.95 – µ)/σ) = 0.15
(1.95 – µ)/σ =-1.036433
µ – 1.036433σ = 1.95 —-(2)
Using the graphic calculator,
µ = 2.03277= 2.03 ( 3 s.f.)
σ = 0.0798699 = 0.0799 ( 3.s.f)
mean mass of seat = 2.03 kg
standard deviation of seat = 0.0799 kg
(b)
Let the mass of the leg be L kg.
L ~ N(1.2, 0.022 )
P(L>1.21) = 0.308537
Expected number of legs with mass more than 1.21 kg = 0.308537(500) = 154.26 =154 (3 s.f.)
(c) Let C = S + L1 + L2 + L3
C ~ N(5.63277, 0.0075792)
P(5.6<C<5.7)= 0.4267085 = 0.427 ( 3 s.f.)
(d) Let W= 0.91S + L1 + L2 + L3
W~N(5.4498207, 0.0064826)
P(W < 5.6) = 0.969456 =0.969 (3 s.f.)
(e)
Let H be the diameter of the seat in mm.
H ~N(31, 0.42)
Let D be the diameter of the legs in mm.
D~ N(30.7, 0.32 )
In order for leg to fit into hole, H > D or H-D > 0
In order for leg to be fitted into hole without padding, H-D < 0.8
H-D ~N(0.3, 0.25)
P(0<H-D < 0.8) = 0.56709
P(all 3 legs fit without need for padding) = (0.56709)3 = 0.18237 = 0.182 ( 3s.f.)
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