(part 2)

b.

• 1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values.
• The tails of the graph of the normal distribution each have an area of 0.40.
• Find k₁, the 40^th percentile, and k₂, the 60^th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution.
• k₁ = invNorm(0.40,5.85,0.24) = 5.79 cm
• k₂ = invNorm(0.60,5.85,0.24) = 5.91 cm

So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm.