Help please. (12 points)
1 answer:
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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 ~
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 .
Answer:
Step-by-step explanation:
Probability (P) is calculated as
P =
The first required outcome is a red sweet from a total of 3 + 5 = 8
P( red) =
There are now 2 red left and a count of 7, since 1 has been eaten, thus
P( second red ) =
P( red and red ) = ×
=
=