Question

For this problem, you must use Excel to perform the necessary calculations. Below are the formulas and equations you will need to use.

If it is appropriate to use the binomial distribution, use the formula =binom.dist(number_s,trials,probability,cumulative) to calculate the probability. In this formula, number_s is the number of successful trials, trials is the total number of trials, probability is the probability for a single trial expressed as a decimal, and cumulative should be set as false. This will report the binomial distribution probability for a single outcome. You are tasked with reporting a cumulative probability.

If it is appropriate to use the normal approximation of the binomial distribution, calculate Z using the correct equation from the option below, and then use either the formula =(1-(norm.dist(Z,0,1,TRUE))) when calculating a right-tailed probability or the formula =(norm.dist(Z,0,1,TRUE)) for a left-tailed probability.

Right-tailed tests find probabilities for X ≥ observed while left-tailed tests find probabilities for X ≤ observed

Observednp Observednp 2 PrIX 2 Observed] - PrZ> or Pr[X s Observed] PrZ< np(1-p)

QUESTION: In Europe, 53% of the flowers of the Rewardless Orchid, Dactylorhiza sambucina, are yellow, whereas the remaining flowers are purple. For this problem, only use the normal approximation where it is appropriate. Use the binomial distribution where the normal approximation is inappropriate.

1. If we took a random sample of a single individual from this population, what is the probability, reported as a percent, that it would be purple?  

2. Determine the probability, reported as a decimal rounded to two decimal places, that if we took a random sample of 6 flowers, at least 4 would be purple.

3. If we took a random sample of 260 individuals, what is the probability, reported as a decimal rounded to four decimal places, that 150 or more of the orchids are purple?

Answer 1

1:

The probability, reported as a percent, that it would be purple is 0.53.

2:

Here we have

n= 6 and p = 0.53

Since np = 3.18 is less than 5 so we normal approximation is not appropriate.

The probability, reported as a decimal rounded to two decimal places, that if we took a random sample of 6 flowers, at least 4 would be purple is

Formual used: =1 - binom.dist(3,6,0.53,true)

Answer: 0.4015

3:

Here we have

n=260 and p=0.53

Using normal approximation, X has approximately normal distribution with mean and SD as follows:

$\mu=np=260\cdot 0.53=137.8$

$\sigma=\sqrt{np(1-p)}=\sqrt{260\cdot 0.53\cdot 0.47}=8.0477$

The z-score for X = 150 -0.5 = 149.5 is

$z=\frac{149.5-137.8}{8.0477}=1.45$

The required probability use:

P(X >= 150) = P(z > 1.45)=0.073529

Formual used: =(1-(norm.dist(1.45,0,1,TRUE)))