Porisma

Worked solution

Combine geometric heights with a weighted class mean

A class has 1616 boys with AP heights and 1010 girls with GP heights. The shortest boy and girl are each 150.0150.0 cm. The tallest boy is 180.0180.0 cm, and the fourth-shortest girl is 157.5157.5 cm.

(i) Show the girls’ common ratio is 1.051/31.05^{1/3} and find the tallest girl’s height to 1 d.p. marks

(ii) Find how many girls are taller than 164.0164.0 cm. marks

(iii) Find the girls’ mean height to 1 d.p. marks

(iv) Find the class mean to 1 d.p. marks

[2013/NYJC/Promo/9]

CJC · Review problems · Question 19

Answer

r=1.051/3r=1.05^{1/3}; tallest girl 173.6173.6 cm; 44 girls above 164.0164.0 cm; girls’ mean 161.6161.6 cm and class mean 163.7163.7 cm.

(i) Calibrate the GP and use the tenth term

150r3=157.5150r^3=157.5 gives r3=1.05r^3=1.05, hence r=1.051/3r=1.05^{1/3}. The tallest girl has height 150r9=150(1.05)3=173.64375≈173.6150r^9=150(1.05)^3=173.64375\approx173.6 cm.

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(ii) Find the first qualifying index and count inclusively

The height of girl n in ascending order is 150(1.05)(n−1)/3150(1.05)^{(n-1)/3}.

nHeight minus 164.0 cm (approximately)
6−1.29
71.38
84.09

Thus indices 7,8,9,107,8,9,10 qualify, giving 10−7+1=410-7+1=4 girls.

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(iii) Use the geometric sum before dividing by ten

The girls’ mean is 150(r10−1)/[10(r−1)]≈161.566150(r^{10}-1)/[10(r-1)]\approx161.566 cm, hence 161.6161.6 cm to 1 d.p. Keep its unrounded value for the overall mean.

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(iv) Weight the two group means by their sizes

The boys’ AP mean is (150+180)/2=165.0(150+180)/2=165.0 cm. The class mean is [16(165.0)+10(unrounded girls’ mean)]/26≈163.7[16(165.0)+10(\text{unrounded girls’ mean})]/26\approx163.7 cm to 1 d.p.

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Common mistakes

The class mean is not the unweighted mean of the boys’ and girls’ means.

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