In Europe, 53% of flowers of the rewardless orchid, Dactylorhiza sambucina, are yellow, whereas the remaining flowers are purple (Gigord et al. 2001). For this problem you may use the normal approximation only if it is appropriate to do so.a. If we took a random sample of a single individual from this population, what is the probability that it would be purple?b. If we took a random sample of five individuals, what is the probability that at least three are yellow?c. If we took many samples of n=5 individuals, what is the expected standard deviation of the sampling distribution for the proportion of yellow flowers?d. If we took a random sample of 263 individuals, what is the probability that no more than 150 are yellow?

a) 0.47

b) 0.5561

c) Standard deviation in the sample proportion of the various samples that are yellow = 0.223

Standard deviation in the number of yellow flowers per sample = 1.12

d) 0.9049

Step-by-step explanation:

In Europe, 53% of the flowers of the rewardless orchid, Dactylorhiza sambucina, are yellow, whereas the remaining flowers are purple.

P(Y) = 0.53

P(P) = P(Y') = 1 - 0.53 = 0.47

a) Probability that a randomly picked flower is purple = P(P) = 1 - 0.53 = 0.47

b) Probability that at least 3 out of 5 flowers are yellow.

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 5 flowers

x = Number of successes required = at least 3 flowers

p = probability of success = probability of a yellow flower = 0.53

q = probability of failure = 1 - p = 0.47

P(X≥3) = P(X=3) + P(X=4) + P(X=5) = 0.328869293 + 0.1854263035 + 0.0418195493 = 0.5561151458 = 0.5561

c) If we took many samples of n=5 individuals, what is the expected standard deviation of the sampling distribution for the proportion of yellow flowers?

Standard deviation in the sample proportion of the various samples that are yellow

= √[p(1-p)/n] = √(0.53×0.47/5) = 0.223

Standard deviation in the number of yellow flowers per sample

= n × √[p(1-p)/n] = √[np(1-p)]

n × √[p(1-p)/n] = 5 × 0.223 = 1.12

OR

√[np(1-p)] = √(5×0.53×0.47) = 1.12

d) If we took a random sample of 263 individuals, what is the probability that no more than 150 are yellow?

Mean = np = 263 × 0.53 = 139.39

Standard deviation of the sample mean = √[np(1-p)] = √[263 × 0.53 × 0.47] = 8.094

probability that no more than 150 are yellow = P(x ≤ 150)

Converting 150 into z-scores,

z = (x - μ)/σ = (150 - 139.39)/8.094 = 1.31

P(x ≤ 150) = P(z ≤ 1.31) = 0.9049

Hope this Helps!!!