Question

High Fructose Corn Syrup (HFCS) is a sweetener in food products that is linked to obesity and type II diabetes. The mean annual consumption in the United States in 2008 of HFCS was 60 lbs with a standard deviation of 20 lbs. Assume the population follows a Normal Distribution.

a. Find the probability a randomly selected American consumes more than 50 lbs of HFCS per year.
b. Find the probability a randomly selected American consumes between 30 and 90 lbs of HFCS per year.
c. Find the 80th percentile of annual consumption of HFCS.
d. In a sample of 40 Americans how many would you expect to consume more than 50 pounds of HFCS per year.
e. Between what two numbers would you expect to contain 95% of Americans HFCS annual consumption?
f. Find the quartile and Interquartile range for this population.
g. A teenager who loves soda consumes 105 lbs of HFCS per year. Is this result unusual?

Answer 1

Explained below.

Explanation:

X = annual consumption in the United States in 2008 of HFCS

`X\sim N(60, 20^(2))`

(a)

Find the probability a randomly selected American consumes more than 50 lbs of HFCS per year.

`P(X>50)=P((X-\mu)/(\sigma)>(50-60)/(20))=P(Z>-0.50)=P(Z<0.50)=0.6915`

P (X > 50) = 0.6915.

(b)

Find the probability a randomly selected American consumes between 30 and 90 lbs of HFCS per year.

`P(30<X<90)=P((30-60)/(20)<(X-\mu)/(\sigma)<(90-60)/(20))\\\\=P(-1.5<Z<1.5)\\\\=0.93319-0.06681\\\\=0.86638\\\\\approx 0.8664`

P (30 < X < 90) = 0.8664.

(c)

Find the 80th percentile of annual consumption of HFCS.

P (X < x) = 0.80

⇒ P (Z < z) = 0.80

⇒ z = 0.84

`z=(x-\mu)/(\sigma)\\\\0.84=(x-60)/(20)\\\\x=60+(20* 0.84}\\\\x=76.8`

80th percentile = 76.8.

(d)

In a sample of 40 Americans how many would you expect to consume more than 50 pounds of HFCS per year.

P (X > 50) = 0.6915

Number of American who consume more than 50 lbs = 40 × 0.6915

= 27.66

≈ 28

Expected number = 28.

(e)

Between what two numbers would you expect to contain 95% of Americans HFCS annual consumption?

According to the Empirical rule, 95% of the normally distributed data lies within 2 standard deviations of mean.

`P(\mu-2\sigma<X<\mu+2\sigma)=0.95\\\\P(60-2\cdot20<X<60+2\cdot20)=0.95\\\\P(20<X<100)=0.95`

Range = 20 < X < 100.

(f)

Find the quartile and Interquartile range for this population.

1st quartile: Q₁

P (X < Q₁) = 0.25

⇒ P (Z < z) = 0.25

⇒ z = -0.67

`z=(Q_(1)-\mu)/(\sigma)\\\\-0.67=(Q_(1)-60)/(20)\\\\Q_(1)=60-(20* 0.67)\\\\Q_(1)=46.6`

3rd quartile: Q₃

P (X < Q₃) = 0.75

⇒ P (Z < z) = 0.75

⇒ z = 0.67

`z=(Q_(3)-\mu)/(\sigma)\\\\0.67=(Q_(3)-60)/(20)\\\\Q_(3)=60+(20* 0.67)\\\\Q_(3)=73.4`

Inter quartile range:

`IQR=Q_(3)-Q_(1)=73.4-46.6=26.8`

(g)

Compute the z-score for x = 105 lbs as follows:

`z=(Q_(3)-\mu)/(\sigma)\\\\z=(105-60)/(20)\\\\z=2.25`

Z-scores greater than +2.00 or less than -2.00 are considered as unusual.

Thus, the result unusual.