F a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8%, can we use the Normal approximation to the Binomial to find the probability of at least 50 obese individuals in our sample?

Probability of at least 50 obese individuals in our sample is 0.92364 .

Step-by-step explanation:

We are given that a random sample of 300 adults is taken from the state of Colorado, where the rate of obesity is 19.8% .

Let X = Number of obese individuals

Firstly, X ~

For approximating binomial distribution into normal distribution, firstly we have to calculate and .

Mean of Normal distribution, = n * p = 300 * 0.198 = 59.4

Variance of Normal distribution, = n * p * (1-p) = 300 *0.198 *0.802 = 47.64

So, now X ~ N()

The standard normal z score distribution is given by;

                     Z = ~ N(0,1)

So, probability of at least 50 obese individuals in our sample = P(X >= 50)

  P(X >= 50) = P(X > 49.5)  {using continuity correction}

  P(X > 49.5) = P( > ) = P(Z > -1.43) = P(Z < 1.43) = 0.92364

Therefore, required probability is 0.92364 .

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

ou would have 12 of each