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Solution — 2025 Question 10 Normal Distribution

(a)

Let R,G and Y be the random variables representing the masses of red, green and yellow peppers respectively

R∼N(230,352)G∼N(225,302)Y∼N(220,322) R1+R2∼N(2(230),2(35)2)R1+R2∼N(460,2450) (G+Y)∼N(225+220,302+322)(G+Y)∼N(445,1924) R1+R2−(G+Y)∼N(460−445,2450+1924)R1+R2−(G+Y)∼N(15,4374)
P(R1+R2>G+Y)=P(R1+R2−(G+Y)>0)=0.58971=0.590∎ (to 3 s.f.)

finding \mathrm{P}\left(R1+ R2 - (G+Y) > 0\right)

(b)
1.2Y∼N(1.2(220),1.22(32)2)1.2Y∼N(264,1474.6) R−1.2Y∼N(230−264,352+1474.6)R−1.2Y∼N(−34,2699.6)
P(R>1.2Y)=P(R−1.2Y≥0)=0.25643=0.256∎ (to 3 s.f.)

finding \mathrm{P}\left(R - 1.2Y \geq 0\right)

(c)
P(R<220)=0.38755

finding \mathrm{P}\left(R < 220\right)

P(G<220)=0.43382

finding \mathrm{P}\left(G < 220\right)

P(Y<220)=0.5

finding \mathrm{P}\left(Y < 220\right)

P(yellow | mass less than 220)=P(yellow ∩ mass less than 220)P( mass less than 220)=0.2(0.5)0.5(0.38755)+0.3(0.43382)+0.2(0.5)=0.23589=0.236∎ (to 3 s.f.)
(d)
P(G>220)=0.56618

finding \mathrm{P}\left(G > 220\right)

Let X represent the number of green peppers in a bag of 20 peppers that has mass at least 220 grams

X∼B(20,0.56618)

P(X≥15)=1−P(X≤14)=1−0.92661=0.073394=0.0734∎ (to 3 s.f.)

probability of X \geq 15

Previous 2024 Q12

Parts

  • Part (a)
  • Part (b)
  • Part (c)
  • Part (d)