Question

The distribution of the amount of turkey individual Americans eat in a year is approximately normal with an average of 16.7 pounds and a standard deviation of 3.8 pounds. 1) You ate 11.5 pounds of turkey last year. What percentile does that put you in

Answer 1

Answer:

82.88%

Step-by-step explanation:

Given that:

Mean (μ) = 16.7 pounds

Standard deviation (σ) = 3.8 pounds

Number of pounds eaten = 11.5 = x

P(11.5 ≥ x ≤11.5)

P(x ≤ 11.5) :

Zscore = (x - μ) / σ

Zscore = (11.5 - 16.7) / 3.8

Zscore = - 5. 2 / 3.8

Zscore = −1.368421

P(Z ≤ - 1.3684) = 0.085593 (Z probability calculator)

P(x ≥ 11.5) ;

Zscore = (x - μ) / σ

Zscore = (11.5 - 16.7) / 3.8

Zscore = - 5. 2 / 3.8

Zscore = −1.368421

P(Z ≥ - 1.3684) = 0.91441 (Z probability calculator)

P(Z ≥ - 1.3684) - P(Z ≤ - 1.3684)

0.91441 - 0.085593 = 0.828817

0.828817 * 100% = 82.88%