Determine the values associated with the 20th and 80th percentile, assuming the data follow the Student T distribution.

Given

The five sample values are:

X₁ = 101.0 cm

X₂ = 106.6 cm

X₃ = 109.1 cm

X₄ = 112.1 cm

X₅ = 115.5 cm

Question(s)

1. Determine the sample mean, XX.

Answer: X=101.0cm+106.6cm+109.1cm+112.1cm+115.5cm5X=101.0cm+106.6cm+109.1cm+112.1cm+115.5cm5= 109 cm

2. Determine the sample variance, s².

Answer: s2=(101.0cm108.8600cm)2+(106.6cm108.8600cm)2+(109.1cm108.8600cm)2+(112.1cm108.8600cm)2+(115.5cm108.8600cm)251s2=(101.0cm-108.8600cm)2+(106.6cm-108.8600cm)2+(109.1cm-108.8600cm)2+(112.1cm-108.8600cm)2+(115.5cm-108.8600cm)25-1= 30.4 cm²

3. Determine the sample standard deviation, s.

Answer: s=30.38300cm2s=30.38300cm2= 5.51 cm

4. Calculate the Degrees of Freedom.

Answer: =51=5-1= 4.0 (Unitless)

5. Using the Student T Distribution, determine the number of standard deviations from the sample mean to the 80th (or 20th) percentile. The combined probability (both tails) is 40 %.

(Answer Length = 6)

6. Estimate the value associated with the 20th percentile.

7. Estimate the value associated with the 80th percentile.