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

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%

the best buy, is the 4- kilogram bag of lollipops for $18.40.

step-by-step explanation:

i found my answer by dividing the cost of each bag by their respective weights.

14.88 divided by 2 = 7.44

17.29 divided by 3 = 5.77

18.4 divided by 4 = 4.6

24.4 divided by 5 = 4.88

by this, i found that the 4 kilo bag is the best buy.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.