Question

26. Kona, a 15-year-old boy, is 72 inches tall. According to the CDC, the average height for boys at this age is 67.00 inches with a standard deviation of 3.19 inches. a. Calculate Kona's z-score. b. What percentage of boys this age are shorter than Kona? c. What percentage of boys this age are as tall as or taller than Kona? d. If lan is in the 30m percentile for height, as a 15 year old boy, how tall is he? e. If statisticians define means that occur less than 5% of the time as "special" or rare, what would you say about Kona's height?

Answer 1

μ = 67, σ = 3.19

(a) x = 72

z = (x – μ)/σ = (72 – 67)/3.19 = 1.5674

(b) P(x < 72) = P(z < 1.5674) = 0.9415 (94.15%)

(c) P(x ≥ 72) = P(z > 1.5674) = 0.0585 (5.85%)

(d) z- score for 30^th percentile is (z- score such that the area to its left is 0.30) z = -0.5244

x = μ + z * σ = 67 – 0.5244 * 3.19 = 65.33 inches

(e) z- score for 5% margin (z- score such that area to its right is 0.05) = 1.645

Since Kona’s z- score < 1.645, Kona’s height can’t be considered as unusual.