According to genetic theory, plants of a particular species have a 25% chance of being red-flowering, independently of other plants. Find the normal approximation to the chance that among 10,000 plants of this species, more than 2400 are red-flowering.
estimate approximate standard deviation to be 2400/4=600
mu=2500, standard deviation 600, above 2400
My calculator says... 0.5662
http://davidmlane.com/hyperstat/z_table.html
To find the normal approximation to the chance that among 10,000 plants of this species, more than 2400 are red-flowering, we can use the central limit theorem.
The central limit theorem tells us that when a large sample size is used, the distribution of sample proportions becomes approximately normal, regardless of the shape of the population distribution.
In this case, we are dealing with a large sample size of 10,000 plants. The proportion of red-flowering plants for each plant is 25% or 0.25.
To apply the normal approximation, we can consider this as a binomial distribution problem. Let X be the number of red-flowering plants out of 10,000. X follows a binomial distribution with n = 10,000 and p = 0.25.
Now, we want to find the probability that more than 2400 plants are red-flowering, which can be expressed as P(X > 2400).
To use the normal approximation, we calculate the mean (μ) and standard deviation (σ) of the binomial distribution:
μ = n * p = 10,000 * 0.25 = 2500
σ = sqrt(n * p * (1 - p)) = sqrt(10,000 * 0.25 * 0.75) = 43.30 (approximately)
Now, we can standardize the binomial distribution using the z-score formula:
z = (x - μ) / σ
where x is the number of red-flowering plants we are interested in.
In this case, we want to find P(X > 2400), which is equivalent to finding P(X ≥ 2401) since the binomial distribution is discrete. Hence, we calculate the z-score for x = 2401:
z = (2401 - 2500) / 43.30 = -2.29 (approximately)
Now, we need to find the probability of z being greater than -2.29. This can be found using the standard normal distribution table or a statistics calculator.
Looking up the z-score of -2.29 in the standard normal distribution table, we find that the probability is approximately 0.011.
Therefore, the normal approximation to the chance that among 10,000 plants of this species, more than 2400 are red-flowering is approximately 0.011 or 1.1%.