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, 0.5^{2})

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, 1.1^{2} x 0.5^{2})

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, σ^{2} )

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)

C_{1} +C_{2} + B ~ N(20 x 1.1 +20 x 1.1 + 12.320 , 1.1^{2} x 0.5^{2}+ 1.1^{2} x 0.5^{2} + 0.47489^{2}) = N(56.320, 0.830520)

P(C_{1} +C_{2} + 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, 1.5^{2})

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, 2^{2})

Let T = B_{1} + B_{2} + B_{3} + B_{4}

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, 1.3^{2}(1.5^{2}) + 2^{2}) = N(140, 7.8025)

P(Y>k) = 0.9

k = 136.420 = 136 ( 3.s.f)

(vi) Let C = Y_{1} + Y_{2} + Y_{3} + Y_{4}

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, 4^{2})

Let the mass of black balls be B grams. B~ N(55, 2^{2})

(i) Let A = W_{1} + W_{2} + W_{3} + W_{4}

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 = B_{1} + B_{2} + B_{3} + W_{1} + W_{2}

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, 0.9^{2})

Let mass of methane molecule be X grams.

X~ 0.7W + 0.9B_{1} + 0.9B_{2} + 0.9B_{3} + 0.9B_{4} + R_{1} + R_{2} + R_{3} + R_{4}

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, 1.2^{2})

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, 3^{2})

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, 6^{2})

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, σ^{2} )

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, 0.02^{2} )

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 + L_{1} + L_{2} + L_{3}

C ~ N(5.63277, 0.0075792)

P(5.6<C<5.7)= 0.4267085 = 0.427 ( 3 s.f.)

(d) Let W= 0.91S + L_{1} + L_{2} + L_{3}

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, 0.4^{2})

Let D be the diameter of the legs in mm.

D~ N(30.7, 0.3^{2} )

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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