Porisma

Worked solution

Keep the instructed rounded model consistent with later probabilities

State the parameters of every normal distribution used. Grapefruits and papayas are sold by mass. The table gives mean mass, standard deviation (SD) and price in dollars per gram.

FruitMean (g)SD (g)Price / g
Grapefruitμ150.007
Papaya450200.008

(i) Given 65% of grapefruits weigh more than 300 g, show that μ=306 to the nearest whole number. marks

Take μ=306 for the remaining parts.

(ii) Find the probability that 3 random grapefruits have total mass below 950 g. marks

(iii) Ten grapefruits are randomly chosen. Find the probability that more than half weigh over 300 g. marks

(iv) Find the probability that the total mass of 3 random grapefruits differs from that of 2 random papayas by at most 100 g. marks

(v) Find the probability that 4 grapefruits cost at least 9 dollars. marks

(vi) A promotion offers one grapefruit and one papaya for a fixed price. Find the least price a dollars giving at least 90% probability of not making a loss. marks

[2018/EJC/MYE/11]

CJC · Review problems · Question 9

Answer

(i)μ≈306 g; (ii)0.891; (iii)0.763 under μ=306, source 0.751 using p=0.65; (iv)0.983; (v)0.0198; (vi)5.99 dollars to cents.

(i) Infer the original mean from its lower-tail quantile

Let X∼N(μ,152)X\sim N(\mu,15^2) and Y∼N(450,202)Y\sim N(450,20^2) be masses in grams. P(X>300)=0.65P(X>300)=0.65 gives P(X≤300)=0.35P(X\le300)=0.35, so P(Z≤(300−μ)/15)=0.35P(Z\le(300-\mu)/15)=0.35. Hence (300−μ)/15=−0.38532(300-\mu)/15=-0.38532 and μ≈305.78, giving 306 to the nearest whole number. The source’s standardisation line uses X in place of Z, and labels 305.78 “3 s.f.”;305.78 actually has 5 significant figures. The required whole-number result is 306.

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(ii) Sum three independent grapefruits

Using the instructed X∼N(306,152)X\sim N(306,15^2) and independently selected fruit, S=X1+X2+X3∼N(3(306),3(152))=N(918,675)S=X_1+X_2+X_3\sim N(3(306),3(15^2))=N(918,675). Thus P(S<950)=0.89097≈0.891P(S<950)=0.89097\approx0.891 to 3 significant figures. The source labels this event P(X<950) after defining the sum; S here avoids that notational ambiguity.

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(iii) Distinguish the printed binomial answer from the rounded model

The source uses W∼B(10,0.65)W\sim B(10,0.65) and P(W>5)=1−P(W≤5)=0.751496≈0.751P(W>5)=1-P(W\le5)=0.751496\approx0.751. This retains the original 65% probability. Under the explicit instruction to take μ=306, however, p=P(X>300)=Φ(0.4)≈0.655421742p=P(X>300)=\Phi(0.4)\approx0.655421742, so W∼B(10,p)W\sim B(10,p) gives P(W>5)≈0.763267838≈0.763P(W>5)\approx0.763267838\approx0.763. The two inputs are inconsistent after rounding the mean; the source 0.751 is labelled rather than silently substituted for the instructed-model result.

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(iv) Use the absolute mass difference

D=X1+X2+X3−(Y1+Y2)∼N(18,1475)D=X_1+X_2+X_3-(Y_1+Y_2)\sim N(18,1475). The parameters come from mean 918−900 and variance 675+800. Therefore P(∣D∣≤100)=P(−100≤D≤100)=0.98256≈0.983P(|D|\le100)=P(-100\le D\le100)=0.98256\approx0.983 to 3 significant figures. Absolute-value bars are retained from the source image.

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(v) Model four-fruit selling price

T=0.007(X1+X2+X3+X4)∼N(0.007(4)(306),0.0072(4)(152))=N(8.568,0.0441)T=0.007(X_1+X_2+X_3+X_4)\sim N(0.007(4)(306),0.007^2(4)(15^2))=N(8.568,0.0441). Thus P(T≥9)=0.019836≈0.0198P(T\ge9)=0.019836\approx0.0198 to 3 significant figures.

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(vi) Model the ordinary price of the bundle

S=0.007X+0.008Y∼N(0.007(306)+0.008(450),0.0072(152)+0.0082(202))=N(5.742,0.036625)S=0.007X+0.008Y\sim N(0.007(306)+0.008(450),0.007^2(15^2)+0.008^2(20^2))=N(5.742,0.036625). Following the source, “not making a loss” means the fixed promotional price is at least the ordinary mass-based selling value S; no purchase costs are supplied.

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(vi) Set the lower-tail guarantee and round upward to cents

P(S≤a)≥0.90P(S\le a)\ge0.90 requires a≥5.9873a\ge5.9873 dollars. The least cent-denominated price is 5.99 dollars, as in the source. Without restricting prices to cents, the lower bound is the unrounded 90 th percentile.

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Common mistakes

The source’s 0.65 binomial parameter conflicts with the instruction to use mean 306. Also distinguish a promotional revenue guarantee from an unspecified profit-cost model.

Solution notes

All six asks/marks, complete mass/price table and explicit parameter instruction, all original normal/binomial models/decimal stages and price quantile retained. Source X/Z and sum-event notation clarified;305.78 significant-figure label qualified. Source part (iii)0.751 retained alongside 0.763 derived from instructed mean 306; independence assumption explicit. Promotional loss interpreted as in source, and cent-price rounding clarified.

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