Porisma

Worked solution

Distinguish two children from twice one child

In a country, men’s weights are normal with mean 70 kg and variance 18 kg², and children’s weights are normal with mean 30 kg and variance 9 kg². The weights of men and children are independent. Find the probability that

(i) a randomly selected man weighs more than two randomly selected children combined,

(ii) a randomly selected man weighs more than twice one randomly selected child. A certain HDB lift has maximum load 340 kg:

(iii) find the probability it is overloaded by 3 men and 4 children;

(iv) find the smallest a such that the probability that the average weight of three randomly chosen men exceeds a kg is less than 0.7.

CJC · Lecture · Example 8

Define the observations and independence model

Let M∼N(70,18)M\sim N(70,18) and C∼N(30,9)C\sim N(30,9) in kg. Use separate indexed variables for different people. The source calculations assume independently selected individuals, including the two children and three men; cross-group independence alone would not establish those within-group assumptions.

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(i) Two separately selected children

M−C1−C2∼N(70−30−30,18+9+9)=N(10,36)M-C_1-C_2\sim N(70-30-30,18+9+9)=N(10,36). Thus P(M>C1+C2)=P(M−C1−C2>0)=0.952P(M>C_1+C_2)=P(M-C_1-C_2>0)=0.952 to three significant figures. The Teacher fills this distribution and probability in the Student working space.

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(ii) Twice one selected child

M−2C∼N(70−2(30),18+22(9))=N(10,54)M-2C\sim N(70-2(30),18+2^2(9))=N(10,54). Hence P(M>2C)=P(M−2C>0)=0.913P(M>2C)=P(M-2C>0)=0.913 to three significant figures. This variance 54 differs from 36 in(i).

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(iii) The lift overload event

M1+M2+M3+C1+C2+C3+C4∼N(3(70)+4(30),3(18)+4(9))=N(330,90)M_1+M_2+M_3+C_1+C_2+C_3+C_4\sim N(3(70)+4(30),3(18)+4(9))=N(330,90). The overload probability is P(M1+M2+M3+C1+C2+C3+C4>340)=0.146P(M_1+M_2+M_3+C_1+C_2+C_3+C_4>340)=0.146 to three significant figures.

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(iv) Obtain the strict cutoff

Put S=M1+M2+M3∼N(3(70),3(18))=N(210,54)S=M_1+M_2+M_3\sim N(3(70),3(18))=N(210,54). Then P(S/3>a)<0.7P(S/3>a)<0.7 is P(S>3a)<0.7P(S>3a)<0.7. First assume equality as in the Teacher sketch: P(S>3a)=0.7P(S>3a)=0.7, giving 3a=206.146458969…3a=206.146458969\ldots. Since the right-tail probability strictly decreases as its cutoff increases, the required inequality is 3a>206.146458969…3a>206.146458969\ldots, or a>68.715486323…a>68.715486323\ldots. This monotonicity explains the Teacher’s lecturer note to demonstrate the inequality.

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(iv) Qualify the requested smallest value

The Teacher gives a≥68.8a\ge68.8. This is the allowable set on a one-decimal grid, whose smallest member is 68.8 kg; the source prompt does not explicitly specify that grid. Among unrestricted real values there is no smallest a, only the infimum 68.715486323…kg, excluded by the strict inequality. The native Student extraction contains a hidden a>68.7a>68.7 line while its rendered working space is blank;68.7 itself does not satisfy the probability requirement. Keep the exact inequality rather than rounding its boundary down.

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Compare your sketch

Expand diagram

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Teacher equality sketch reconstructed: S, the sum of three independent men’s weights, is normal with mean 210 and variance 54. A cutoff 3a near 206.146 lies left of the mean and leaves right-tail area 0.7. For the requested strict probability below 0.7, move that cutoff to the right; hence a>68.715486… and 68.8 is the smallest one-decimal value.
Teacher equality sketch reconstructed: S, the sum of three independent men’s weights, is normal with mean 210 and variance 54. A cutoff 3a near 206.146 lies left of the mean and leaves right-tail area 0.7. For the requested strict probability below 0.7, move that cutoff to the right; hence a>68.715486… and 68.8 is the smallest one-decimal value.
Teacher equality sketch reconstructed: S, the sum of three independent men’s weights, is normal with mean 210 and variance 54. A cutoff 3a near 206.146 lies left of the mean and leaves right-tail area 0.7. For the requested strict probability below 0.7, move that cutoff to the right; hence a>68.715486… and 68.8 is the smallest one-decimal value.

Common mistakes

The second parameter is variance. A strict probability inequality has an open real cutoff; a rounded displayed number is not automatically the smallest admissible real value.

Read the hint