Porisma

Worked solution

Separate contents, packaging and independent packet sums

The contents of a packet of cereal A have a normal mass distribution with mean 625 g and standard deviation 15 g. Packaging has an independent normal distribution with mean 25 g and standard deviation 3 g. Give answers to 3 significant figures.

(a) Find the probability that one packet of A has total mass exceeding 630 g.

(b) Find the probability that the contents of 4 random packets of A exceed 2450 g.

Cereal B contents have normal mass with mean 465 g and standard deviation 10 g. Find the probability that the contents of 4 random packets of B outweigh the contents of 3 random packets of A.

[J1987/II/8]

CJC · Discussion · Question 10

Answer

(a)0.904; (b)0.952; final comparison 0.324 (3 s.f.).

Model contents and packaging separately

Let A be the mass in grams of cereal A contents and P its packaging: A∼N(625,152)A\sim N(625,15^2), P∼N(25,32)P\sim N(25,3^2), independently. Distinct randomly selected packets are also treated as independent in the source sum models.

How did this part go?

(a) Contents plus random packaging

A+P∼N(625+25,152+32)=N(650,234)A+P\sim N(625+25,15^2+3^2)=N(650,234). Thus P(A+P>630)=0.904P(A+P>630)=0.904 to 3 significant figures.

How did this part go?

(b) Contents only: four independent packets

A1+A2+A3+A4∼N(4(625),4(225))=N(2500,900)A_1+A_2+A_3+A_4\sim N(4(625),4(225))=N(2500,900). Hence P(A1+A2+A3+A4>2450)=0.952P(A_1+A_2+A_3+A_4>2450)=0.952 to 3 significant figures. Packaging is excluded from this ask.

How did this part go?

Compare four B packets with three A packets

Let B∼N(465,102)B\sim N(465,10^2). Then B1+B2+B3+B4−A1−A2−A3∼N(4(465)−3(625),4(100)+3(225))=N(−15,1075)B_1+B_2+B_3+B_4-A_1-A_2-A_3\sim N(4(465)-3(625),4(100)+3(225))=N(-15,1075).

How did this part go?

Rewrite the event as a positive difference

P(B1+B2+B3+B4>A1+A2+A3)=P(B1+B2+B3+B4−A1−A2−A3>0)=0.324P(B_1+B_2+B_3+B_4>A_1+A_2+A_3)=P(B_1+B_2+B_3+B_4-A_1-A_2-A_3>0)=0.324 to 3 significant figures.

How did this part go?

Common mistakes

Random packaging contributes variance; a fixed shift would not. Four independent packets are not 4 times one packet.

Solution notes

All three asks, every source component/sum/difference distribution with expanded mean/variance terms, event rewrite and probability retained. Distinct-packet independence assumption explicit; no marks printed.

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