Statistics — Exercise 13.3

Use the direct method: represent each class by its midpoint xᵢ, multiply by fᵢ, and divide the product total by the frequency total. These are grouped-data estimates.

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{9320}{68}\approx 137.058824\]

Mean ≈ 137.058824 units.

N=68, so N/2=34. The median class is 125–145. Its lower boundary is 125, preceding cumulative frequency 22, class frequency 20 and width 20.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=125+\frac{12}{20}\times 20\approx 137\]

Median ≈ 137 units.

The modal class is 125–145, with largest frequency 20. Use l=125, h=20, f₁=20, f₀=13, f₂=14.

\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=125+\frac{7}{13}\times 20\approx 135.769231\]

Modal estimate ≈ 135.769231 units.

The three central values are close, with mode < median < mean. Grouping produces estimates rather than individual exact observations.

Known frequencies sum to 5+20+15+5=45, so x+y=15. The median class is 20–30: l=20, h=10, f=20, preceding cumulative frequency=5+x and N/2=30.

\[28.5=20+\frac{30-(5+x)}{20}10\]

\[8.5=\frac{25-x}{2}\;\Rightarrow\;17=25-x\]

Thus x=8 and y=15−8=7.

The “below” counts are cumulative: 2,6,24,45,78,89,92,98,100. Subtract adjacent totals to get the frequencies 2,4,18,21,33,11,3,6,2. The first class begins at the eligibility age 18.

N=100, so N/2=50. The median class is 35–40. Its lower boundary is 35, preceding cumulative frequency 45, class frequency 33 and width 5.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=35+\frac{5}{33}\times 5\approx 35.757576\]

Median ≈ 35.757576 years.

Because lengths were rounded to the nearest millimetre, use boundaries 117.5–126.5,126.5–135.5,…,171.5–180.5. Each width is 9 mm; do not use the unadjusted lower limit 145.

N=40, so N/2=20. The median class is 144.5–153.5. Its lower boundary is 144.5, preceding cumulative frequency 17, class frequency 12 and width 9.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=144.5+\frac{3}{12}\times 9\approx 146.75\]

Median ≈ 146.75 mm.

N=400, so N/2=200. The median class is 3000–3500. Its lower boundary is 3000, preceding cumulative frequency 130, class frequency 86 and width 500.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=3000+\frac{70}{86}\times 500\approx 3406.976744\]

Median ≈ 3406.976744 hours.

Use the direct method: represent each class by its midpoint xᵢ, multiply by fᵢ, and divide the product total by the frequency total. These are grouped-data estimates.

\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{832}{100}\approx 8.32\]

Mean ≈ 8.32 letters.

N=100, so N/2=50. The median class is 7–10. Its lower boundary is 7, preceding cumulative frequency 36, class frequency 40 and width 3.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=7+\frac{14}{40}\times 3\approx 8.05\]

Median ≈ 8.05 letters.

The modal class is 7–10, with largest frequency 40. Use l=7, h=3, f₁=40, f₀=30, f₂=16.

\[\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}h=7+\frac{10}{34}\times 3\approx 7.882353\]

Modal estimate ≈ 7.882353 letters.

Fractional answers estimate the centre of grouped integer counts; they are not literal lengths of an individual surname.

N=30, so N/2=15. The median class is 55–60. Its lower boundary is 55, preceding cumulative frequency 13, class frequency 6 and width 5.

\[\text{Median}=l+\frac{N/2-c_f}{f}h=55+\frac{2}{6}\times 5\approx 56.666667\]

Median ≈ 56.666667 kg.