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 .

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: