Respuesta :
Answer:
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!!!
The probability that no more than 150 out of 263 are yellow is 0.9049.
What is Z-score?
A Z-score helps us to understand how far is the data from the mean. It is a measure of how many times the data is above or below the mean. It is given by the formula,
[tex]Z = \dfrac{X- \mu}{\sigma}[/tex]
Where Z is the Z-score,
X is the data point,
μ is the mean and σ is the standard variable.
A.) As it is given that the probability of a flower being yellow is 53%, therefore, the probability that a flower is not yellow(Purple) can be written as,
[tex]\text{Probability of Yellow}+\text{Probability of Purple} = 1\\\\\text{Probability of Purple} = 1 -0.53\\\\\text{Probability of Purple} = 0.47[/tex]
B.) The probability that at least three are yellow out of the five can be written as,
[tex]P(X\geq 3) = P(X=3)+P(X=4)+P(X=5)[/tex]
[tex]= (0.53^3 \times 0.47^2)+(0.53^4 \times 0.47^1)+(0.53^5 \times 0.47^0)\\\\=0.0329+0.037+0.04182\\\\=0.1117[/tex]
C.) The expected standard deviation of the sampling distribution for the proportion of yellow flowers.
Standard deviation is the sample proportion of the various samples that are yellow
[tex]=\sqrt{\dfrac{p(1-p)}{n}} =\sqrt{\dfrac{0.53(1-0.53)}{5}}=0.223[/tex]
Standard deviation in the number of yellow flowers per sample
[tex]=n\sqrt{\dfrac{p(1-p)}{n}} =5\times \sqrt{\dfrac{0.53(1-0.53)}{5}}=5 \times 0.223=1.12[/tex]
D.) If we took a random sample of 263 individuals, then the probability that no more than 150 are yellow,
[tex]\rm Mean, \mu = np = 263\times 0.53 = 139.39\\\\Standard\ deviation, \sigma= \sqrt{np(1-p)} = \sqrt{263\times0.53 \times 0.47} = 8.094[/tex]
Using the Z-table,
[tex]\begin{aligned}P(X\leq 150)&= P(z\leq \dfrac{150-\mu}{\sigma})\\&= P(z\leq \dfrac{150-139.39}{8.094})\\&=0.9049\end{aligned}[/tex]
Hence, the probability that no more than 150 are yellow is 0.9049.
Learn more about Z-score:
https://brainly.com/question/13299273
