If a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 20%, to easily find the probability that we will find at least 50 people in the sample who are obese we use the normal .

Step-by-step explanation:

Since the the rate of obesity is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = number of obese adults.

µ = mean

σ = standard deviation

From the information given,

p = 20/100 = 0.2

n = 300

µ = np = 300 × 0.2 = 60

q = 1 - p = 1 - 0.2 = 0.8

σ = √npq = √300 × 0.2 × 0.8 = 6.93

The probability that we will find at least 50 people in the sample who are obese is expressed as

P(x ≥ 50) = 1 - P(x < 50)

For x = 50

z = (50 - 60)/6.93 = - 1.44

Looking at the normal distribution table, the probability corresponding to the z score is 0.075

P(x ≥ 50) = 1 - 0.075 = 0.925

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2