Question

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?

Answer 1

Answer:

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 ~ $Binom(n=300,p=0.198)$

For approximating binomial distribution into normal distribution, firstly we have to calculate $\mu$ and $\sigma^{2}$ .

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

Variance of Normal distribution,$\sigma^{2}$ = n * p * (1-p) = 300 *0.198 *0.802 = 47.64

So, now X ~ N($\mu = 59.4 , \sigma^{2} = 47.64$)

The standard normal z score distribution is given by;

                     Z = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ > $\frac{49.5 - 59.4}{6.9}$ ) = P(Z > -1.43) = P(Z < 1.43) = 0.92364

Therefore, required probability is 0.92364 .