Respuesta :
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 ~ [tex]Binom(n=300,p=0.198)[/tex]
For approximating binomial distribution into normal distribution, firstly we have to calculate [tex]\mu[/tex] and [tex]\sigma^{2}[/tex] .
Mean of Normal distribution, [tex]\mu[/tex] = n * p = 300 * 0.198 = 59.4
Variance of Normal distribution,[tex]\sigma^{2}[/tex] = n * p * (1-p) = 300 *0.198 *0.802 = 47.64
So, now X ~ N([tex]\mu = 59.4 , \sigma^{2} = 47.64[/tex])
The standard normal z score distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{49.5 - 59.4}{6.9}[/tex] ) = P(Z > -1.43) = P(Z < 1.43) = 0.92364
Therefore, required probability is 0.92364 .
